# 3.1: Graphs of Quadratic Functions - Mathematics We are searching data for your request:

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A Quadratic Function is any function defined by a polynomial whose greatest exponent is two. That means it can be written in the form (f(x)=ax^2+bx+c), with the restrictions that (a) canNOT be zero, and (a), (b), and (c) are real numbers.

The graph of any quadratic function is a U-shaped curve called a parabola. There are certain key features that are important to recognize that can be seen on a graph and that can be calculated from an equation.

 1. The orientation of a parabola is that it either opens up or opens down2. The vertex is the lowest or highest point on the graph3. The axis of symmetry is the vertical line that goes through the vertex, dividing the parabola into two equal parts. If (h) is the (x)-coordinate of the vertex, then the equation for the axis of symmetry is (x=h).4. The maximum or minimum value of a parabola is the (y)-coordinate of the vertex.5. The (y)-intercept is the point at which the parabola crosses the (y)-axis.The (x)-intercepts are the points at which the parabola crosses the (x)-axis.6. The domain of a parabola is all real numbers, ( (-infty, infty) ).The range of a parabola starts or ends with the value of the (y)-coordinate of the vertex. Figure (PageIndex{1}): Graph illustrating features of a parabola .

Example (PageIndex{1}): Identify Features of a Parabola from a graph

Determine features of th parabola illustrated below. Solution.Orientation: opens upVertex: ((3,1))Axis of symmetry: (x=3)Minimum Value: (y=1)(y)-intercept: ((0,7))(x)-intercept: noneDomain: ( (-infty, infty) )Range: ([1, infty))

## Forms of a Quadratic Function

There are two important forms of a quadratic function

A quadratic function is a parabolic function of degree two. The graph of a quadratic function is a parabola.

• The general form of a quadratic function is (f(x)=ax^2+bx+c) with real number parameters (a), (b), and (c) and (a{ eq}0).
• The standard form of a quadratic function is (f(x)=a(x−h)^2+k) with real number parameters (a), (h), and (k) and (a{ eq}0). Standard form is also known as vertex form.

The graph of (f(x)=a(x−h)^2+k) is a graph of (y=x^2) that has undergone some transformations. The order in which these transformations are performed is reviewed in the steps below.

1. If (h>0), the graph shifts toward the right (h) units and if (h<0), the graph shifts to the left (h) units.
2. If (a<0), the graph has been reflected over the (x)-axis.
3. The magnitude of (a) indicates the stretch of the graph. If (|a|>1) there is a vertical stretch because the point associated with a particular (x)-value shifts farther from the (x)-axis, and the graph appears to become narrower. But if (|a|<1) there is a vertical compression because the point associated with a particular (x)-value shifts closer to the (x)-axis, and the graph appears to become wider.
4. If (k>0), the graph shifts upward (k) units, whereas if (k<0), the graph shifts downward (k) units.

In conclusion, it is clear that the vertex is located at the point ((h, k)). Therefore, if the equation is in standard form, it is very easy to determine the location of the vertex.

The location of the vertex if the equation is in general form can be determined by using the fact that these two forms of a quadratic equation describe the same function. So a formula for (h) can be found by expanding out the standard form and setting it equal to the general form.

[egin{align*} a(x−h)^2+k &= ax^2+bx+c [4pt] ax^2−2ahx+(ah^2+k)&=ax^2+bx+c end{align*} ]

For the linear terms to be equal, the coefficients must be equal.

[–2ah=b ext{, so } h=−dfrac{b}{2a}. onumber]

This is the (x)-coordinate of the vertex. To find what the corresponding (y)-coordinate of the vertex is, evaluate the function when (x=h). In other words, (f(h)=k).

Features of the graph of a quadratic function depend on the parameter values (a), (b), (c) or (a), (h), (k) used in its equation. A summary of how features of the parabola for a quadratic function can be obtained is summarized below.

HowTo: Find Features of a Parabola Given a Quadratic Equation

• Orientation
• (a > 0), the parabola opens up ( stackrel {+ ::: +} {igcup} )
• (a < 0 ), the parabola opens down ( stackrel{- ::: -}{igcap } )
• The vertex is located at ((h,k)).
• If the function is in general form, calculate (h) and (k): (h=dfrac{-b}{2a}, qquad k=f(h)=f(dfrac{−b}{2a}).)
• The axis of symmetry, (x=h) is the ( underline { extrm{equation}} ) of the vertical line through the vertex.
• The maximum or minimum value depends on the (y)-coordinate, (k), of the vertex and the orientation of the parabola
• If the parabola opens up, ((a > 0)), the vertex is the lowest point on the graph so the graph has a minimum value (k).
• If the parabola opens down, ((a < 0)), the vertex is the highest point on the graph so the graph has a maximum value (k)
• The domain is always ( mathbb{R}) or ( (-infty, infty) ).
The range depends on the (y)-coordinate, (k), of the vertex and the orientation of the parabola
• If the parabola opens up, ((a > 0)), the vertex is the lowest point on the graph so the range of the function is ( [k,infty) ).
• If the parabola opens down, ((a < 0)), the vertex is the highest point on the graph so the range of the function is ((-infty, k] ).
• Intercepts are the points where the parabola crosses the axes.
• The (x)-intercepts are the points ((s, 0)), where (s) is a real solution to (f(x)=0)
• The (y)-intercept is the point ((0, f(0))).

### Orientation

When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.

Example (PageIndex{2}): Find the orientation of a parabola

Determine whether each parabola opens upward or downward:

 a. (f(x)=-3 x^{2}+2 x-4)a. Solution:Find the value of (a). ( quad ) Since the (a) is negative, the parabola will open downward. b. (f(x)=6 (x+1)^{2}-11)b. Solution:Find the value of (a).(f(x) = a(x - h)^2 + k)(f(x)=6(x+1)^2-11)(a=6)( quad ) Since the (a) is positive, the parabola will open upward.

Try It (PageIndex{2})

Determine whether the graph of each function is a parabola that opens upward or downward:

### Vertex and Axis of Symmetry

When given a quadratic in standard form (f(x)=a(x−h)^2+k), the vertex and axis of symmetry is easily found once the parameters (h) and (k) have been identified. (Notice the sign on (h)!!) The vertex is ((h, k)) and the axis of symmetry is the vertical line ( x=h).

When given a quadratic in general form: (f(x)=ax^2+bx+c), more computation is required. After identifying parameters (a) and (b), calculate (h=–dfrac{b}{2a}), and then evaluate the function for that value of (x) to find the corresponding (y) coordinate for that point on the graph: ( k=f(h) ). The axis of symmetry is the vertical line through the vertex, so its equation is ( x=h).

Example (PageIndex{3a}): Find the Vertex from the General Form of the Quadratic Equation

For the graph of (f(x)=3x^{2}-6 x+2) find:

1. the axis of symmetry
2. the vertex

Solution:

( egin{array}{llc}
ext{a.} & ext{Identify the equation parameters} & a=3, b=-6, c=2
& ext{The axis of symmetry is the vertical line } x=-frac{b}{2 a} &
& ext{Substitute the values }a ext{ and } b ext{ into the formula} & x=-frac{-6}{2 cdot 3}
& ext{Simplify.} & x=1
&& ext{The axis of symmetry is the line } x=1
end{array} )

( egin{array}{llc}
ext{b.} & ext{The vertex is a point on the line of symmetry, so} & ext{ The } x ext{ coordinate of the vertex is } x=1
& ext{The }y ext{ coordinate will be }f(1) & f(1)=3({color{red}{1}})^2-6({color{red}{1}})+2
& ext{Simplify} & f(1) = 3-6+2
& ext{The result is the }y ext{ coordinate of the vertex.} & f(1)=-1
&& ext{The vertex is } (1, -1)
end{array} )

Example (PageIndex{3b}): Find the Vertex from the Standard Form of the Quadratic Equation

For the graph of (f(x)=6(x-3)^{2}+4) find:

1. the axis of symmetry
2. the vertex

Solution:

( egin{array}{lll}
ext{a.} & ext{Identify the equation parameters} & a=6, h=3, k=4
& ext{The axis of symmetry is the vertical line } x=h &
& ext{Substitute.} & ext{The axis of symmetry is the line } x=3
\
ext{b.} & ext{Use the equation parameters} & a=6, h=3, k=4
& ext{The vertex is the point } (h, k) & ext{The vertex is the point } (3,4)
end{array} )

Try It (PageIndex{3})

For the following quadratic functions find a. the axis of symmetry and b. the vertex

### Minimum or Maximum Value of a Quadratic Function

 Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic function.The (y)-coordinate of the vertex of the graph of a quadratic function isthe minimum value of the quadratic equation if the parabola opens upward.the maximum value of the quadratic equation if the parabola opens downward. Example (PageIndex{4})

Find the minimum or maximum value of the quadratic function (f(x)=x^{2}+2 x-8).

Solution:

( egin{array}{llc}
ext{Identify the equation parameters} & a=1, b=2, c=-8
ext{Since }a ext{ is positive, the parabola opens upward. } & ext{The quadratic equation has a minimum.}
ext{State the formula for the axis of symmetry } & x=-frac{b}{2 a}
ext{Substitute.} & x=-frac{2}{2 cdot 1}
ext{Simplify.} & x=-1 qquad qquad ext{The axis of symmetry is the line } x=-1
ext{The }y ext{ coordinate will be }f(-1) & f(-1)=({color{red}{-1}})^2-2({color{red}{-1}})-8
ext{Simplify} & f(-1) = 1-2-8
ext{The result is the }y ext{ coordinate of the vertex.} & f(-1)=-9 qquad ext{The vertex is } (-1, -9)
end{array} )

Since the parabola has a minimum, the (y)-coordinate of the vertex is the minimum (y)-value of the quadratic equation. The minimum value of the quadratic is (-9) and it occurs when (x=-1).

Try It (PageIndex{4})

Find the maximum or minimum value of the quadratic function

1. (f(x)=x^{2}-8 x+12).
2. (f(x)=-4(x-2)^{2}+5).

a. The minimum value of the quadratic function is (−4) and it occurs when (x=4).

b. The maximum value of the quadratic function is (5) and it occurs when (x=2).

### Domain and Range

Any number can be the input value ( (x) ) to a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all (y)-values greater than or equal to the (y)-coordinate at the turning point or less than or equal to the (y)-coordinate at the turning point, depending on whether the parabola opens up or down. Given a quadratic function, find the domain and range.

1. The domain of any quadratic function is always ( mathbb{R}) or ( (-infty, infty) ).
2. Determine the maximum or minimum value of the parabola, (k)
1. If the function is in the form (f(x)=a(x−h)^2+k), then the value of (k) is readily visible as one of the parameters.
2. If the function is in the form (f(x)=ax^2+bx+c), the vertex must be determined and the value for (k) is the (y) coordinate of the vertex.
3. Determine whether (a) is positive or negative.
1. If (a) is positive, the parabola has a minimum value of (k) and the range of the function is ( [k,infty) ).
2. If (a) is negative, the parabola has a maximum value of (k) and the range of the function is ((-infty, k] ).

Example (PageIndex{5}): Find the Domain and Range of a Quadratic Function

Find the domain and range of (f(x)=−5x^2+9x−1).

Solution

As with any quadratic function, the domain is all real numbers.

Because (a) is negative, the parabola opens downward and has a maximum value.
The maximum value must be determined. Begin by finding the (x)-value of the vertex.

( h=−dfrac{b}{2a} =−dfrac{9}{2(-5)}=dfrac{9}{10} )

The maximum value is given by (f(h)).

(f(frac{9}{10})=−5(frac{9}{10})^2+9(frac{9}{10})-1 = frac{61}{20})

The range is (f(x){leq}frac{61}{20}), or (left(−infty,frac{61}{20} ight]).

Try It (PageIndex{5})

Find the domain and range of (f(x)=2Big(x−frac{4}{7}Big)^2+frac{8}{11}).

The domain is all real numbers. The range is (f(x){geq}frac{8}{11}), or (left[frac{8}{11},infty ight)).

### Intercepts

The (y)-intercept is the point where the graph crosses the (y) axis. All points on the (y)-axis have an (x) coordinate of zero, so the (y)-intercept of a quadratic is found by evaluating the function (f(0)).

The (x)-intercepts are the points where the graph crosses the (x)-axis. All points on the (x)-axis have a (y) coordinate of zero, so the (x)-intercept of a quadratic can be found by solving the equation (f(x)=0). Notice in Figure (PageIndex{6}) that the number of (x)-intercepts can vary depending upon the location of the graph.  Given a quadratic function (f(x)), find the (y)- and (x)-intercepts.

1. Evaluate (f(0)) to find the (y)-intercept.
2. Solve the quadratic equation (f(x)=0) to find the (x)-intercepts.

Example (PageIndex{6}): Finding the (y)- and (x)-intercepts of a General Form Quadratic

Find the (y)- and (x)-intercepts of the quadratic (f(x)=3x^2+5x−2).

Solution

Find the (y)-intercept by evaluating (f(0)).

( f(0)=3(0)^2+5(0)−2 =−2 )

So the (y)-intercept is at ((0,−2)).

For the (x)-intercepts, find all solutions of (f(x)=0).

(0=3x^2+5x−2)

In this case, the quadratic can be factored easily, providing the simplest method for solution. Typically, quadratics in general form, like this example, are usually solved using factoring, or failing that, using the quadratic formula or complete the square.

(0=(3x−1)(x+2))

[egin{align*} 0&=3x−1 & 0&=x+2 x&= frac{1}{3} & ext{or} ;;;;;;;; x&=−2 end{align*} ]

So the (x)-intercepts are at ((frac{1}{3},0)) and ((−2,0)).

Example (PageIndex{7}):

Find the (y)- and (x)-intercepts of the quadratic ( f(x)=x^2+x+2).

Solution

Find the (y)-intercept by evaluating (f(0)).

( f(0)=(0)^2+(0)+2 =2 )

So the (y)-intercept is at ((0,−2)).

For the (x)-intercepts, find all solutions of (f(x)=0) or ( x^2+x+2 = 0). Clearly this does not factor, so employ the quadratic formula.

The quadratic formula: (x=frac{−b{pm}sqrt{b^2−4ac}}{2a}) and for this equation, (a=1), (b=1), and (c=2). Substituting these values into the formula produces

[egin{align*} x&=dfrac{−1{pm}sqrt{1^2−4⋅1⋅(2)}}{2⋅1} &=dfrac{−1{pm}sqrt{1−8}}{2} &=dfrac{−1{pm}sqrt{−7}}{2} qquad =dfrac{−1{pm}isqrt{7}}{2} onumber end{align*}]

Since the solutions are imaginary, there are no (x)-intercepts.

Example (PageIndex{8}): Find the (y)- and (x)-intercepts of a Standard Form Quadratic

Find the (y)- and (x)-intercepts of the quadratic (f(x)=-2(x+3)^2+5).

Solution

Find the (y)-intercept by evaluating (f(0)). Notice that the quantity inside the parentheses (((0+3)=(3))) is evaluated FIRST!!!

( egin{align*}
f(0) &=-2(0+3)^2+5
&=-2(3)^2+5\
&=-2(9)+5\
&=-18+5 = -13\
end{align*} )

So the (y)-intercept is at ((0,−13)).

For the (x)-intercepts, find all solutions of (f(x)=0). Solving a quadratic equation given in standard form, like in this example, is most efficiently accomplished by using the Square Root Property

( egin{array}{c}
0 =-2(x+3)^2+5
2(x+3)^2 =5
(x+3)^2 =dfrac{5}{2}
x+3 =pm sqrt{ dfrac{5}{2}}
x = -3 pm sqrt{ frac{5}{2}}
end{array} )

So the (x)-intercepts are at ( (-3+sqrt{2.5}, 0) ) and ( (-3-sqrt{2.5}, 0) ).

Try It (PageIndex{8})

Find the (y)- and (x)-intercepts for the function (g(x)=13+x^2−6x).

(y)-intercept at ((0, 13)), No (x)-intercepts

The previous sections detailed the pieces needed in order to graph a quadratic function. A summary of these steps and examples appear below. Graph a quadratic function in the form (f(x)=ax^2+bx+c)

1. Determine whether the parabola opens upward ( (a > 0) ) or downward ( (a < 0) ).
2. Find the equation of the axis of symmetry, ( x = h ) where (h=– frac{b}{2a} ).
3. Find the vertex, ( (h, k) ), where ( k = f(h) ).
4. Find the (y)-intercept, ( f(0) ). Find the point symmetric to the (y)-intercept across the axis of symmetry.
5. Find the (x)-intercepts. (Set (f(x)=0) and solve for x using factoring, QF or CTS). Find additional points if needed.
6. Graph the parabola.

Example (PageIndex{9}) How to Graph a General Form Quadratic Function Using Properties

Graph (f(x)=x^{2}-6x+8) by using its properties.

Solution:

 Step 1: Determine whether the parabola opens upward or downward. Look (a) in the equation (f(x)=x^{2}-6x+8)Since (a) is positive, the parabola opens up. (f(x)=x^{2}-6x+8)(color{red}{a=1,; b=-6, ;c=8})The parabola opens upward. Step 2: Find the axis of symmetry. (f(x)=x^{2}-6x+8)The axis of symmetry is the line (x=-frac{b}{2 a}). Axis of Symmetry(x=-frac{b}{2 a})(x=-frac{(-6)}{2 cdot 1})(x=3)The axis of symmetry is the line (x=3). Step 3: Find the vertex. The vertex is on the axis of symmetry. Substitute (x=3) into the function. Vertex(f(x)=x^{2}-6x+8)(f(3)=(color{red}{3}color{black}{)}^{2}-6(color{red}{3}color{black}{)}+8)(f(3)=-1)The vertex is ((3,-1)). Step 4: Find the (y)-intercept. Find the point symmetric to the (y)-intercept across the axis of symmetry. Find (f(0)).Use the axis of symmetry to find a point symmetric to the (y)-intercept. The (y)-intercept is (3) units left of the axis of symmetry, (x=3). A point (3) units to the right of the axis of symmetry has (x=6). (y)-intercept(f(x)=x^{2}-6 x+8)(f(0)=(color{red}{0}color{black}{)}^{2}-6(color{red}{0}color{black}{)}+8)(f(0)=8)The (y)-intercept is ((0,8)).Point symmetric to (y)-intercept:The point is ((6,8)). Step 5: Find the (x)-intercepts. Find additional points if needed. Solve (f(x)=0).Solve this quadratic equation by factoring. (x)-intercepts(f(x)=x^{2}-6 x+8)(color{red}{0}color{black}{=}x^{2}-6x+8)(color{red}{0}color{black}{=}(x-2)(x-4))(x=2) or (x=4)The (x)-intercepts are ((2,0)) and ((4,0)). Step 6: Graph the parabola. We graph the vertex, intercepts, and the point symmetric to the (y)-intercept. We connect these (5) points to sketch the parabola. Try It (PageIndex{9})

Graph the following quadratic functions by using its properties.

 a. (f(x)=x^{2}+2x-8)a. Answeropens up, vertex: ((-1,-9)), axis: (x=-1),intercepts: ((0, -8), (-4,0), (2, 0) ), symm. pt: ((-2, -8) ) b. (f(x)=x^{2}-8x+12)b. Answeropens up, vertex: ((4, -4)), axis: (x=4),intercepts: ( (0, 12), (2, 0), (6, 0) ), symm.pt: ((8,12))  Graph a quadratic function in the form (f(x)=a(x-h)^2+k)

1. Determine whether the parabola opens upward ( (a > 0) ) or downward ( (a < 0) ).
2. Find the equation of the axis of symmetry, ( x = h ).
3. Find the vertex, ( (h, k) ).
4. Find the (y)-intercept, ( (f(0) ). Find the point symmetric to the (y)-intercept across the axis of symmetry.
5. Find the (x)-intercepts. (Use the square root property to solve (a(x-h)^2+h=0). Find additional points if needed.
6. Graph the parabola.

Example (PageIndex{10}): How to Graph a Vertex Form Quadratic Using Properties

Graph the function (f(x)=2(x+1)^{2}+3) by using its properties

Solution:

 Step 1: Determine whether the parabola opens upward or downward. Identify the constants (a, h, k).Since (a=2), the parabola opens upward. (a=2,; h=-1,; k=3)The parabola opens upward. Step 2: Find the axis of symmetry. The axis of symmetry is (x=h). The axis of symmetry is the line (x=-1). Step 3: Find the vertex. The vertex is ((h,k)). The vertex is ((-1,3)). Step 4: Find the (y)-intercept. Find the point symmetric to the (y)-intercept across the axis of symmetry. Find the (y)-intercept by finding (f(0)).The (y)-intercept is (1) units right of the axis of symmetry, (x=-1). A point (1) units to the left of the axis of symmetry has (x=-2). (f(0)=2(0+1)^2+3=2(1)+3=5)The (y)-intercept is ( (0,5) ).Point symmetric to (y)-intercept is ( (-2,5) ) Step 5: Find the (x)-intercepts. Find additional points if needed. Solve (f(x)=0).Use the Square Root Property.This equation has imaginary solutions, so there are no (x)-intercepts ( egin{array} {c}2(x+1)^{2}+3=0 2(x+1)^{2}=-3 (x+1)^{2}=-3/2 x+1 = pm sqrt(-3/2) x=-1 pm isqrt(1.5)end{array} )No (x) intercepts Step 6: Graph the parabola. Graph the vertex, intercepts, and the point symmetric to the (y)-intercept. Connect these points to sketch the parabola.Two more points:( f(1) = 2(1+1)^2+3 = 2(2^2)+3=11)Therefore, ( (1, 11) ) is on the graph.By symmetry, the point ( (-3, 11) ) is also on the graph Try It (PageIndex{10})

Graph the following functions using properties

 a. (f(x)=3(x-1)^{2}+2)a. Answeropens up, vertex: ((1,2)), axis: (x=1),intercepts: ((0, 5) ), symm. pt: ((2,5) ) b. (f(x)=-2(x-2)^{2}+1)b. Answeropens down, vertex: ((2,1)), axis: (x=2),intercepts: ((0, -7), (approx 1.3,0), (approx 2.7, 0) ), symm. pt: ((4, -7) ) ## Rewrite Quadratics into Standard Form

As the above examples illustrate, it is often easier to graph a quadratic equation that is in standard form, rather than in general form. This is particularly true when trying to find (x)-intercepts for equations that don't easily factor. There are two different approaches for transforming an equation in general form into an equation in standard (or vertex) form. One method uses the formulas for (h) and (k). The other method uses Complete the Square. Both will be illustrated below.

### Formula method Rewrite (y=ax^2+bx+c) into vertex form - formula method.

1. Identify constants ( a) and (b).
2. Substitute ( a) and (b) into the formula: (h=−frac{b}{2a}).
3. Substitute (x=h) into the general form of the quadratic function to find (k).
4. Rewrite the quadratic in standard form using (h) and (k). The standard form of the function is (f(x)=a(x−h)^2+k ).

Example (PageIndex{11}): Formula method of rewriting into standard form

Rewrite the quadratic function (f(x)=2x^2+4x−4) into standard form.

Solution

Step 1. Values of the parameters in the general form are (a=2), (b=4), and (c=-4).
The parameter (a) is the same in both forms of the function, so (a=2).

Step 2. Solve for (h).

[egin{align*} h&=−dfrac{b}{2a} &=−dfrac{4}{2(2)} =−1 end{align*}]

Step 3. Use the value found for (h) to find (k).

[egin{align*} k&=f(h)=f(−1) &=2(−1)^2+4(−1)−4 =−6 end{align*}]

Step 4. The standard form of the function is:

[egin{align*} f(x)&=a(x−h)^2+k f(x)&=2(x+1)^2−6 end{align*}]

Try It (PageIndex{11})

Find the standard form for the function (g(x)=13+x^2−6x).

(g(x)=(x-3)^2 +4 )

### Complete the square method

Another way of transforming (f(x)=ax^{2}+bx+c) into the form (f(x)=a(x−h)^{2}+k) is by completing the square. This form is known as the vertex form or standard form. This approach will also be used when circles are studied.

We must be careful to both add and subtract the number to the SAME side of the function to complete the square. We cannot add the number to both sides as we did when we completed the square with quadratic equations. When we complete the square in a function with a coefficient of (x^{2}) that is not one, we have to factor that coefficient from the (x)-terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the (x)-terms.

Once we get the constant we want to complete the square, we must remember to multiply it by the coefficient that was part of the (x^2) term before we then subtract it. Rewrite (y=ax^2+bx+c) into vertex form - complete the square method.

1. Separate the (x) terms from the constant.
2. If the coefficient of (x^{2}) is not 1, factor it out from the (x^2) and (x) terms.
3. Find the CTS constant needed to complete the square on the (x^2) and (x) terms.
4. Add the CTS constant to the (x^2) and (x) terms and subtract the CTS constant (multiplied by the coefficient of (x^{2}) if not 1)
5. Write the trinomial as a binomial square and combine constants outside the binomial square to arrive at the standard form of the function.

Example (PageIndex{12}): CTS method of rewriting into vertex form

Rewrite (f(x)=−3x^{2}−6x−1) in the (f(x)=a(x−h)^{2}+k) form by completing the square.

Solution:

 Step 1. Separate the (x) terms from the constant. (f(x)=−3x^{2}−6x−1)(f(x)=−3x^{2}−6x qquad qquad−1) Step 2. Factor the coefficient of (x^{2}, -3). (f(x)=−3(x^{2}+2x) qquad qquad−1) Step 3. Prepare to complete the square. (f(x)=−3(x^{2}+2x qquad qquad) −1) Take half of (2) and then square it to complete the square ((frac{1}{2}cdot 2)^{2}=1) (f(x)=-3(x^2+2x + largeBox ) −1+3 largeBox ) Step 4. The constant (1) completes the square in the parentheses, but the parentheses is multiplied by (-3). So we are really adding (-3). We must then add (3) to not change the value of the function. Step 5. Rewrite the trinomial as a square and combine the constants. ( f(x) = -3(x+1)^2+2 ) The function is now in the (f(x)=a(x-h)^{2}+k) form. Try It (PageIndex{12})

Rewrite the following functions in the (f(x)=a(x−h)^{2}+k) form by completing the square.

## Obtain the Equation of a Quadratic Function from a Graph

So far we have started with a function and then found its graph.

Now we are going to reverse the process. Starting with the graph, we will find the function.

HOWTO: Write a quadratic function in a general form

Given a graph of a quadratic function, write the equation of the function in general form.

1. Identify the horizontal shift of the parabola; this value is (h). Identify the vertical shift of the parabola; this value is (k).
2. Substitute the values of the horizontal and vertical shift for (h) and (k). in the function (f(x)=a(x–h)^2+k).
3. Substitute the values of any point, other than the vertex, on the graph of the parabola for (x) and (f(x)).
4. Solve for the stretch factor, (|a|).
5. If the parabola opens up, (a>0). If the parabola opens down, (a<0) since this means the graph was reflected about the (x)-axis.
6. Expand and simplify to write in general form.

Example (PageIndex{13}): Writing the Equation of a Quadratic Function from the Graph

Write an equation for the quadratic function (g) in Figure (PageIndex{13}) as a transformation of (f(x)=x^2), and then expand the formula, and simplify terms to write the equation in general form. Solution

Since it is quadratic, we start with the (g(x)=a(x−h)^{2}+k) form. (Observe the minus sign in front of (h)!) The vertex, ((h,k)), is ((−2,−3)) so (h=−2) and (k=−3). Substituting theses values we obtain (g(x)=a(x+2)^2–3).

Substituting the coordinates of a point on the curve, such as ((0,−1)), we can solve for the stretch factor.

[egin{align*} −1&=a(0+2)^2−3 2&=4a a&=dfrac{1}{2} end{align*}]

In standard form, the algebraic model for this graph is (g(x)=dfrac{1}{2}(x+2)^2–3).

To write this in general polynomial form, we can expand the formula and simplify terms.

[egin{align*} g(x)&=dfrac{1}{2}(x+2)^2−3 &=dfrac{1}{2}(x+2)(x+2)−3 &=dfrac{1}{2}(x^2+4x+4)−3 &=dfrac{1}{2}x^2+2x+2−3 &=dfrac{1}{2}x^2+2x−1 end{align*}]

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

Example (PageIndex{14})

Determine the quadratic function whose graph is shown. Solution:

The vertex, ((h,k)), is ((−2,−1)) so (h=−2) and (k=−1).

To find (a), we use the (y)-intercept, ((0,7)).

So (f(0)=7).

(7=a(0+2)^{2}-1)

Solve for (a).

(egin{array}{l}{7=4 a-1} {8=4 a} {2=a}end{array})

Write the function.

(f(x)=2(x+2)^{2}-1)

Try It (PageIndex{14})

Write the quadratic function in (f(x)=a(x−h)^{2}+k) form whose graph is shown.

 a. a. Answer(f(x)=(x-3)^{2}-4) b. b. Answer(f(x)=(x+3)^{2}-1)

Try It (PageIndex{15})

A coordinate grid has been superimposed over the quadratic path of a basketball in Figure (PageIndex{15}). Find an equation for the path of the ball. Does the shooter make the basket? Figure (PageIndex{15}): Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes.
(credit: modification of work by Dan Meyer)

The path passes through the origin and has vertex at ((−4, 7)), so (h(x)=–frac{7}{16}(x+4)^2+7). To make the shot, (h(−7.5)) would need to be about 4 but (h(–7.5){approx}1.64); he doesn’t make it.

## Key Equations

• general form of a quadratic function: (f(x)=ax^2+bx+c)
• standard form of a quadratic function: (f(x)=a(x−h)^2+k)

## Key Concepts

• A polynomial function of degree two is called a quadratic function.
• The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down.
• The axis of symmetry is the vertical line passing through the vertex. The zeros, or (x)-intercepts, are the points at which the parabola crosses the (x)-axis. The (y)-intercept is the point at which the parabola crosses the (y)-axis.
• Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph.
• The vertex can be found from an equation representing a quadratic function. .
• The domain of a quadratic function is all real numbers. The range varies with the function.
• A quadratic function’s minimum or maximum value is given by the (y)-value of the vertex.
• Some quadratic equations must be solved by using the quadratic formula.

## Glossary

axis of symmetry
a vertical line drawn through the vertex of a parabola around which the parabola is symmetric; it is defined by (x=−frac{b}{2a}).

general form of a quadratic function
the function that describes a parabola, written in the form (f(x)=ax^2+bx+c), where (a,b,) and (c) are real numbers and a≠0.

standard form of a quadratic function
the function that describes a parabola, written in the form (f(x)=a(x−h)^2+k), where ((h, k)) is the vertex.

vertex
the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function

vertex form of a quadratic function
another name for the standard form of a quadratic function

zeros
in a given function, the values of (x) at which (y=0), also called roots

## 3.1: Graphs of Quadratic Functions - Mathematics

So far, we have seen how to graph linear functions, i.e. functions whose graphs are straight lines. In this section, we will discuss a type of function that is in a sense the next step up. Straight lines are created by functions whose highest power of x is 1, and here we will discuss functions whose highest power of x is 2. These functions have graphs that are called parabolas.

In order to get the general form of the quadratic from either of the others, you need only multiply out and simplify, thus this form is also called the expanded form . To get the factored form, you must first have the function in general form, and then factor the expression. And to get the standard form you will want to use the technique of Completing the Square this too requires the general form as a starting point.

• Direction : whether the parabola opens up or down.
• The vertex of the parabola
• The x -intercept(s) (there may be 0, 1, or 2 x -intercepts)
• The y -intercept (there is always one)

These are the four fundamental items of information that we need to know about a parabola in order to sketch it.

There are two basics shapes for the graph of a quadratic function. In most cases you will be able to deduce the direction of the parabola, i.e. whether it opens up or opens down , by applying the following rule for a quadratic in general form:

A particularly important point in the graph of a quadratic is called the vertex . This point is either the maximum or the minimum point of the parabola. If the parabola opens down, the vertex is the maximum, if the parabola opens up, then the vertex is the minimum point. The coordinates of the vertex are most readily seen using the standard form of the quadratic.

For example, in the quadratic function we saw above, the standard form is so the vertex is at the point

Justification for the connection between the formula in standard form and the vertex comes from the graphing techniques we studied earlier. For the quadratic , the vertex is the origin, Subtracting h from x means we have a right horizontal shift by h units if h is positive, or a left horizontal shift by | h | units if h is negative. Adding k to the rest of the expression means we have a vertical shift up by k units if k is positive, or a vertical shift down by | k | units if k is negative. Thus, the new vertex is at ( h, k ) .

If you are given a quadratic function in general form, then to find the vertex you can either rewrite the expression in standard form or else use the following formula.

This formula is derived by rewriting in standard form.

In the example above, this gives a vertex with and y = = = . Thus, the vertex is again shown to be at

The most convenient form of the quadratic to use to find the x- intercepts is the factored form we then set each of the factors equal to 0. In our example above, since the factored form of the function is
the x- intercepts are x = 1 and Notice that the x value of the vertex is half way between these, as we would expect.

Not every quadratic function has 2 x- intercepts. There may be one or even no x- intercepts, as we see below.

 y = x 2 + 6 x + 9 When is there only one x -intercept? This happens when the factored form of the quadratic is a perfect square. Try, for example to factor y = x 2 + 1 When are there no x -intercepts? This will happen when the quadratic is not factorizable via real numbers. That is, we would have to use complex numbers in order to factor it. Try, for example, to factor y = x 2 + 1 and see what happens!

Another option for finding the x -intercepts is the quadratic formula . We use it when the function is given in general form, in this case the x -coordinate of the x -intercepts are given by

Every quadratic has a (single) y -intercept. The reason for this is that the y -intercept is the function value at x = 0, and we can always substitute x = 0 into the quadratic. Thus, the y -intercept of the quadratic function For the other forms of the function, just substitute x = 0 to find the corresponding value of y .

Look at this example of a geometric problem that leads to a quadratic function.

## 3.1: Graphs of Quadratic Functions - Mathematics

(page 2 of 4)

Sections: Introduction, The meaning of the leading coefficient / The vertex, Examples

The general form of a quadratic is " y = ax 2 + bx + c ". For graphing, the leading coefficient " a " indicates how "fat" or how "skinny" the parabola will be.

For | a | > 1 (such as a = 3 or a = &ndash4 ), the parabola will be "skinny", because it grows more quickly (three times as fast or four times as fast, respectively, in the case of our sample values
of a ).

For | a | < 1 (such as a = 1 /3 or a = &ndash1 /4 ), the parabola will be "fat", because it grows more slowly (one-third as fast or one-fourth as fast, respectively, in the examples). Also, if a is negative, then the parabola is upside-down.

You can see these trends when you look at how the curve y = ax 2 moves as " a " changes:

As you can see, as the leading coefficient goes from very negative to slightly negative to zero (not really a quadratic) to slightly positive to very positive, the parabola goes from skinny upside-down to fat upside-down to a straight line (called a "degenerate" parabola) to a fat right-side-up to a skinny right-side-up. Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

There is a simple, if slightly "dumb", way to remember the difference between right-side-up parabolas and upside-down parabolas:

This can be useful information: If, for instance, you have an equation where a is negative, but you're somehow coming up with plot points that make it look like the quadratic is right-side-up, then you will know that you need to go back and check your work, because something is wrong .

Parabolas always have a lowest point (or a highest point, if the parabola is upside-down). This point, where the parabola changes direction, is called the "vertex".

If the quadratic is written in the form y = a(x &ndash h) 2 + k, then the vertex is the point (h, k) . This makes sense, if you think about it. The squared part is always positive (for a right-side-up parabola), unless it's zero. So you'll always have that fixed value k , and then you'll always be adding something to it to make y bigger, unless of course the squared part is zero. So the smallest y can possibly be is y = k , and this smallest value will happen when the squared part, x &ndash h , equals zero. And the squared part is zero when x &ndash h = 0 , or when x = h . The same reasoning works, with k being the largest value and the squared part always subtracting from it, for upside-down parabolas.

(Note: The " a " in the vertex form " y = a(x &ndash h) 2 + k " of the quadratic is the same as the " a " in the common form of the quadratic equation, " y = ax 2 + bx + c ".)

Since the vertex is a useful point, and since you can "read off" the coordinates for the vertex from the vertex form of the quadratic, you can see where the vertex form of the quadratic can be helpful, especially if the vertex isn't one of your T-chart values. However, quadratics are not usually written in vertex form. You can complete the square to convert ax 2 + bx + c to vertex form, but, for finding the vertex, it's simpler to just use a formula . (The vertex formula is derived from the completing-the-square process, just as is the Quadratic Formula. In each case, memorization is probably simpler than completing the square.)

For a given quadratic y = ax 2 + bx + c , the vertex (h, k) is found by computing h = &ndashb /2a , and then evaluating y at h to find k . If you've already learned the Quadratic Formula , you may find it easy to memorize the formula for k , since it is related to both the formula for h and the discriminant in the Quadratic Formula: k = (4ac &ndash b 2 ) / 4a .

To find the vertex, I look at the coefficients a , b , and c . The formula for the vertex gives me:

Then I can find k by evaluating y at h = &ndash1 /6 :

So now I know that the vertex is at ( &ndash1 /6 , &ndash25 /12 ) . Using the formula was helpful, because this point is not one that I was likely to get on my T-chart.

I need additional points for my graph:

Now I can do my graph, and I will label the vertex:

When you write down the vertex in your homework, write down the exact coordinates: " ( &ndash1 /6 , &ndash25 /12 ) ". But for graphing purposes, the decimal approximation of " (&ndash0.2, &ndash2.1) " may be more helpful, since it's easier to locate on the axes.

The only other consideration regarding the vertex is the "axis of symmetry". If you look at a parabola, you'll notice that you could draw a vertical line right up through the middle which would split the parabola into two mirrored halves. This vertical line, right through the vertex, is called the axis of symmetry. If you're asked for the axis, write down the line " x = h" , where h is just the x -coordinate of the vertex. So in the example above, then the axis would be the vertical line x = h = &ndash1 /6.

Helpful note: If your quadratic's x -intercepts happen to be nice neat numbers (so they're relatively easy to work with), a shortcut for finding the axis of symmetry is to note that this vertical line is always exactly between the two x -intercepts. So you can just average the two intercepts to get the location of the axis of symmetry and the x- coordinate of the vertex. However, if you have messy x -intercepts (as in the example above) or if the quadratic doesn't actually cross the x -axis (as you'll see on the next page), then you'll need to use the formula to find the vertex.

## 3.1: Graphs of Quadratic Functions - Mathematics ### Graphing

There are a few pieces of information that you have to put together in order to create a graph of a quadratic function. Before anything, though, we need to learn the standard form of a quadratic and this is it:

For example, what are a,b, and c in the following expression?

Great! That’s the first step. Next little piece of info that you will need is that every time you graph a quadratic, it is going to look like a “u” shape, which is called a parabola. If your first number (your “a”) is positive then it will look like a smiley face parabola (opens up) and if “a” is negative, then your parabola will be a frowny face (opens down).

 (x^2=) (-x^2=) We are almost ready to graph. Two more pieces to go. The first one is something called the axis of symmetry. Symmetry means there is a line that exists that you can fold over to make the graph overlap itself. If you look at the two graphs above, can you see where that line might be? It’s right down the center at the y-axis on both graphs.

This line is going to help us graph. Here’s how you find it:

Axis of Symmetry (AOS) ► (x=frac<-b><2a>)

These are your same a and b as before. Let’s use our above example:

Our AOS is x = -1. We can show this on the graph by drawing a dotted line at x = -1. Your AOS (in this chapter) will always be a vertical line. Almost done! We need one more thing. It’s called the vertex, which is the highest or lowest point depending on whether the parabola opens up or opens down.

To find the vertex, take the value (the “x”) that you found for the AOS and plug it into the equation to find the “y”.

So, your vertex is at the point (-1,-5). If you want to sketch the graph, you now have enough information. If we want it to be very accurate, we need a few more points. Pick some before and after the AOS and plug into your equation.

 (x) (2x^2 + 4x - 3) (y) -3 (2(-3)^2 + 4(-3) - 3) 3 -2 (2(-2)^2 + 4(-2) - 3) -3 -1 (2(-1)^2 + 4(-1) - 3) -5 0 (2(0)^2 + 4(0) - 3) -3 1 (2(1)^2 + 4(1) - 3) 3 ## The effects of (a), (p) and (q) on a parabolic graph

On the same system of axes, plot the following graphs:

Use your sketches of the functions above to complete the following table:

On the same system of axes, plot the following graphs:

1. (y_1 = x^2 + 2)
2. (y_2 = (x - 2)^2 - 1)
3. (y_3 = (x - 1)^2 + 1)
4. (y_4 = (x + 1)^2 + 1)
5. (y_5 = (x + 2)^2 - 1)

Use your sketches of the functions above to complete the following table:

Consider the three functions given below and answer the questions that follow:

What is the value of (a) for (y_2)?

Does (y_1) have a minimum or maximum turning point?

What are the coordinates of the turning point of (y_2)?

Compare the graphs of (y_1) and (y_2). Discuss the similarities and differences.

What is the value of (a) for (y_3)?

Will the graph of (y_3) be narrower or wider than the graph of (y_1)?

Determine the coordinates of the turning point of (y_3).

Compare the graphs of (y_1) and (y_3). Describe any differences.

The effect of the parameters on (y = a(x + p)^2 + q)

The effect of (p) is a horizontal shift because all points are moved the same distance in the same direction (the entire graph slides to the left or to the right).

For (p>0), the graph is shifted to the left by (p) units.

For (p<0), the graph is shifted to the right by (p) units.

The value of (p) also affects whether the turning point is to the left of the (y)-axis (left(p>0 ight)) or to the right of the (y)-axis (left(p<0 ight)). The axis of symmetry is the line (x = -p).

The effect of (q) is a vertical shift. The value of (q) affects whether the turning point of the graph is above the (x)-axis (left(q>0 ight)) or below the (x)-axis (left(q<0 ight)).

The value of (a) affects the shape of the graph. If (a<0), the graph is a “frown” and has a maximum turning point. If (a>0) then the graph is a “smile” and has a minimum turning point. When (a = 0), the graph is a horizontal line (y = q).

## Graphical Solutions of Quadratic Equations

We can solve a quadratic equation by factoring, completing the square, using the quadratic formula or using the graphical method.

Compared to the other methods, the graphical method only gives an estimate to the solution(s).

If the graph of the quadratic function crosses the x-axis at two points then we have two solutions. If the graph touches the x-axis at one point then we have one solution. If the graph does not intersect with the x-axis then the equation has no real solution.

The following diagrams show the three types of solutions that a quadratic equation can have: two solutions, one solution and no real solution. Scroll down the page for more examples and solutions. We also have a quadratic equations calculator that can solve quadratic equations algebraically and graphically.

How to solve quadratic equations graphically using x-intercepts

The following video explains how the quadratic graph can show the number of solutions for the quadratic equation and the values of the solutions.

Examples of how to use the graph of a quadratic function to solve a quadratic equation: Two solutions, one solution and no solution.

1. Use the graph of y = x 2 + x - 6 to solve x 2 + x - 6 = 0
2. Use the graph of y = -x 2 + 4 to solve -x 2 + 4 = 0
3. Use the graph of y = x 2 -2x + 1 to solve x 2 -2x + 1
4. Use the graph of y = x 2 + 1 to solve x 2 + 1

### Quadratic Equation With Two Solutions

We will now graph a quadratic equation that has two solutions. The solutions are given by the two points where the graph intersects the x-axis.

Example:
Solve the equation x 2 + x – 3 = 0 by drawing its graph for –3 ≤ x ≤ 2.

Solution:
Rewrite the quadratic equation x 2 + x – 3 = 0 as the quadratic function y = x 2 + x – 3

Draw the graph for y = x 2 + x – 3 for –3 ≤ x ≤ 2.

 x &ndash3 &ndash2 &ndash1 0 1 2 y 3 &ndash1 &ndash3 &ndash3 &ndash1 3

The solution for the equation x 2 + x – 3 can be obtained by looking at the points where the graph y = x 2 + x – 3 cuts the x-axis (i.e. y = 0).

The graph y = x 2 + x – 3, cuts the x-axis at x 1.3 and x –2.3

So, the solution for the equation x + x –3 is x 1.3 or x –2.3.

Recall that in the quadratic formula, the discriminant b 2 – 4ac is positive when there are two distinct real solutions (or roots).

How to solve quadratic equation by graphing?

It uses the vertex formula to get the vertex which also gives an idea of what values to choose to plot the points. This is an example where the coefficient of x 2 is positive.

Example:
Solve the following quadratic equation by graphing
x 2 - 4x + 3 = 0

Find the roots of a quadratic equation by graphing

This video shows an example of solving quadratic equation by graphing. It uses the vertex formula to get the vertex which also gives an idea of what values to choose to plot the points. This is an example where the coefficient of x 2 is negative.

Example:
Solve the following quadratic equation by graphing
-2x 2 + 4x + 4 = 0

### Quadratic Equation With One Solution Example: By plotting the graph, solve the equation 6x – 9 – x 2 = 0. x 0 1 2 3 4 5 6 y &ndash9 &ndash4 &ndash1 0 &ndash1 &ndash4 &ndash9 Notice that the graph does not cross the x-axis, but touches the x-axis at x = 3. This means that the equation 6x – 9 – x 2 = 0 has one solution (or equal roots) of x = 3. Recall that in the quadratic formula, in such a case where the roots are equal, the discriminant b 2 – 4ac = 0. Quadratic Equation With No Real Solution

Example:
Solve the equation x 2 + 4x + 8 = 0 using the graphical method.

 x &ndash4 &ndash3 &ndash2 &ndash1 0 1 y 8 5 4 5 8 13

Notice that the graph does not cross or touch the x-axis. This means that the equation x 2 + 4x + 8 = 0 does not have any real solution (or roots).

Recall that in the quadratic formula, the discriminant b 2 – 4ac, is negative when there are no real solution (or roots).

Solving Quadratic Equations by Graphing Part 1

This video demonstrates how to solve quadratic equations by graphing.

1. Solve one side of the equation for zero.
2. Change the zero to y or f(x).
3. Graph the function.
4. Read the solutions where the function crosses or touches the x-axis.

Roots, x-intercepts, and zeros are given as synonyms for solutions. Finding roots from a table of values is also demonstrated.

Solving Quadratic Equations by Graphing Part 2

This video shows how to solve quadratic equations using the TI84 and TI83 series of graphing calculators.

Five problems are worked out. The different steps are shown including converting quadratic equations into calculator ready graphable quadratic functions.

The video shows how to examine in graph and table view what the solutions are. The case of having no solutions is shown as well as that of having only one solution. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

#### Solving a Single Variable Equation :

Add 1 to both sides of the equation :
x = 1

#### Parabola, Finding the Vertex :

2.3 Find the Vertex of y = x 2 +x+1

Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola, Ax 2 +Bx+C, the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -0.5000

Plugging into the parabola formula -0.5000 for x we can calculate the y -coordinate :
y = 1.0 * -0.50 * -0.50 + 1.0 * -0.50 + 1.0
or y = 0.750

#### Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x 2 +x+1
Axis of Symmetry (dashed) = <-0.50>
Vertex at = <-0.50, 0.75>
Function has no real roots

#### Solve Quadratic Equation by Completing The Square

2.4 Solving x 2 +x+1 = 0 by Completing The Square .

Subtract 1 from both side of the equation :
x 2 +x = -1

Now the clever bit: Take the coefficient of x , which is 1 , divide by two, giving 1/2 , and finally square it giving 1/4

Add 1/4 to both sides of the equation :
On the right hand side we have :
-1 + 1/4 or, (-1/1)+(1/4)
The common denominator of the two fractions is 4 Adding (-4/4)+(1/4) gives -3/4
So adding to both sides we finally get :
x 2 +x+(1/4) = -3/4

Adding 1/4 has completed the left hand side into a perfect square :
x 2 +x+(1/4) =
(x+(1/2)) • (x+(1/2)) =
(x+(1/2)) 2
Things which are equal to the same thing are also equal to one another. Since
x 2 +x+(1/4) = -3/4 and
x 2 +x+(1/4) = (x+(1/2)) 2
then, according to the law of transitivity,
(x+(1/2)) 2 = -3/4

We'll refer to this Equation as Eq. #2.4.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
(x+(1/2)) 2 is
(x+(1/2)) 2/2 =
(x+(1/2)) 1 =
x+(1/2)

Now, applying the Square Root Principle to Eq. #2.4.1 we get:
x+(1/2) = √ -3/4

Subtract 1/2 from both sides to obtain:
x = -1/2 + √ -3/4
In Math, i is called the imaginary unit. It satisfies i 2 =-1. Both i and -i are the square roots of -1

Since a square root has two values, one positive and the other negative
x 2 + x + 1 = 0
has two solutions:
x = -1/2 + √ 3/4 • i
or
x = -1/2 - √ 3/4 • i

Note that √ 3/4 can be written as
√ 3 / √ 4 which is √ 3 / 2

2.5 Solving x 2 +x+1 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax 2 +Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B 2 -4AC
x = ————————
2A

In our case, A = 1
B = 1
C = 1

Accordingly, B 2 - 4AC =
1 - 4 =
-3

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)

Both i and -i are the square roots of minus 1

Accordingly, √ -3 =
√ 3 • (-1) =
√ 3 • √ -1 =
± √ 3 • i

√ 3 , rounded to 4 decimal digits, is 1.7321
So now we are looking at:
x = ( -1 ± 1.732 i ) / 2

## Use graphing to solve quadratic equations

You know by now how to solve a quadratic equation using factoring. Another way of solving a quadratic equation is to solve it graphically. The roots of a quadratic equation are the x-intercepts of the graph.

Graph the equation. This could either be done by making a table of values as we have done in previous sections or by computer or a graphing calculator. The parabola cross the x-axis at x = -2 and x = 5. These are the roots of the quadratic equation.

We can compare this solution to the one we would get if we were to solve the quadratic equation by factoring as we've done earlier.

$left ( x+2 ight )left ( x-5 ight )=0$

• A quadratic equation has two roots if its graph has two x-intercepts
• A quadratic equation has one root it its graph has one x-intercept
• A quadratic equation has no real solutions if its graph has no x-intercepts.

The old standard form for a parabola was written like any other polynomial, f(x) = ax 2 + bx + c, a &ne 0.

We're going to complete the square and place it into a form where the translations are easily interpreted. This time, instead of dividing through by a, let's factor an a out of the x-terms instead.

Go ahead and take half of the x-coefficient and put it on the next line.

One thing to be careful of here. When you add the b 2 /(4a 2 ), you are really multiplying it by the a that you factored out, so it is really just a b 2 /(4a). This time, instead of adding it to both sides of the equation, add it and subtract it on the same side of the equation.

f(x) = a [ x 2 + (b/a) x + b 2 /(4a 2 ) ] + c - b 2 /(4a)

f(x) = a [ x + (b/2a) ] 2 + (4ac - b 2 )/(4a)

With a couple of substitutions, this can be written in the new standard form.

where h = -b/(2a) and k = (4ac - b 2 ) / (4a)

Do not worry about what k is, but you might want to memorize the value for h.

The x-coordinate of the vertex is -b/(2a). The y-coordinate is what you get when you plug -b/(2a) back into the original function for x.

There are three translations involved here.

• The y-coordinates have been multiplied by a. This is the same a that was in the original problem. If a>0, then the parabola opens up and the vertex is at the bottom. If a<0, then the parabola opens down and the vertex is at the top.
• There has been a horizontal shift. Instead of the x-coordinate of the vertex being at x=0, it is now at x=h, where h=-b/(2a). Since the axis of symmetry passes through the vertex, that means that the axis of symmetry is now x=-b/(2a).
• There has been a vertical shift. The y-coordinate of the vertex is now at y=k. It is not worth your time to memorize the formula for the vertical shift. It isn't that hard, it is -a times the discriminant of the quadratic, but it is easier to find the x-coordinate, and plug that back into the equation to find the y-coordinate.

Unless the coefficients are really nasty (ie, decimals), you may find it quicker to complete the square to find the vertex than to let x=-b/(2a) and then find the y-coordinate.

But do note that the vertex is now at (h,k) instead of (0,0).

## Ways to Find the Roots of a Quadratic Function

### Factorization

The most common way people learn how to determine the the roots of a quadratic function is by factorizing. For a lot of quadratic functions this is the easiest way, but it also might be very difficult to see what to do. We have a quadratic function ax^2 + bx + c, but since we are going to set it equal to zero, we can divide all terms by a if a is not equal to zero. Then we have an equation of the form:

Now we try to find factors s and t such that:

If we succeed we know that x^2 + px + q = 0 is true if and only if (x-s)(x-t) = 0 is true. (x-s)(x-t) = 0 means that either (x-s) = 0 or (x-t)=0. This means that x = s and x = t are both solutions, and hence they are the roots.

If (x-s)(x-t) = x^2 + px + q, then it holds that s*t = q and - s - t = p.

Numerical Example

Then we have to find s and t such that s*t = 15 and - s - t = 8. So if we choose s = -3 and t = -5 we get:

Hence, x = -3 or x = -5. Let&aposs check these values: (-3)^2 +8*-3 +15 = 9 - 24 + 15 = 0 and (-5)^2 + 8*-5 +15 = 25 - 40 + 15 = 0. So indeed these are the roots.

It might however be very difficult to find such a factorization. For example:

Then the roots are 3 - sqrt 2 and 3 + sqrt 2. These are not so easy to find.

### The ABC Formula

Another way to find the roots of a quadratic function. This is an easy method that anyone can use. It is just a formula you can fill in that gives you roots. The formula is as follows for a quadratic function ax^2 + bx + c:

(-b + sqrt(b^2 -4ac))/2a and (-b - sqrt(b^2 -4ac))/2a

This formulas give both roots. When only one root exists both formulas will give the same answer. If no roots exist, then b^2 -4ac will be smaller than zero. Therefore the square root does not exist and there is no answer to the formula. The number b^2 -4ac is called the discriminant.