# 10.7: Vertex Form of a Quadratic Equation

Difficulty Level: At Grade Created by: CK-12
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Practice Vertex Form of a Quadratic Equation

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What if you had a quadratic function like y2=x2+4x\begin{align*}y - 2 = x^2 + 4x\end{align*}? How could you rewrite it in vertex form to find its vertex and intercepts? After completing this Concept, you'll be able to rewrite and graph quadratic equations like this one in vertex form.

### Guidance

Probably one of the best applications of the method of completing the square is using it to rewrite a quadratic function in vertex form. The vertex form of a quadratic function is

yk=a(xh)2\begin{align*}y - k = a(x - h)^2\end{align*}

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point (h,k)\begin{align*}(h, k)\end{align*}.

It is also simple to find the x\begin{align*}x-\end{align*}intercepts from the vertex form: just set y=0\begin{align*}y = 0\end{align*} and take the square root of both sides of the resulting equation.

To find the y\begin{align*}y-\end{align*}intercept, set x=0\begin{align*}x = 0\end{align*} and simplify.

#### Example A

Find the vertex, the x\begin{align*}x-\end{align*}intercepts and the y\begin{align*}y-\end{align*}intercept of the following parabolas:

a) y2=(x1)2\begin{align*}y - 2 = (x - 1)^2\end{align*}

b) y+8=2(x3)2\begin{align*}y + 8 = 2(x - 3)^2\end{align*}

Solution

a) y2=(x1)2\begin{align*}y - 2 = (x - 1)^2\end{align*}

Vertex: (1, 2)

To find the x\begin{align*}x-\end{align*}intercepts,

Set y=0:Take the square root of both sides:22=(x1)2=x1and2=x1\begin{align*}\text{Set} \ y = 0: & & -2 & = (x - 1)^2 \\ \text{Take the square root of both sides}: & & \sqrt{-2} & = x - 1 && \text{and} && -\sqrt{-2} = x - 1\end{align*}

The solutions are not real so there are no x\begin{align*}x-\end{align*}intercepts.

To find the y\begin{align*}y-\end{align*}intercept,

Set x=0:Simplify:y2y2=(1)2=1y=3\begin{align*}\text{Set} \ x = 0: & & y - 2 & = (-1)^2\\ \text{Simplify}: & & y - 2 & = 1 \Rightarrow \underline{y = 3}\end{align*}

b) y+8=2(x3)2\begin{align*}y + 8 = 2(x - 3)^2\end{align*}

Rewrite:Vertex:y(8)=2(x3)2(3,8)\begin{align*}& \text{Rewrite}: & & y - (-8) = 2(x - 3)^2\\ & \text{Vertex}: & & \underline{(3, -8)}\end{align*}

To find the x\begin{align*}x-\end{align*}intercepts,

Set y=0:Divide both sides by 2:Take the square root of both sides:Simplify:844=2(x3)2=(x3)2=x3x=7andand4=x3x=1\begin{align*}\text{Set} \ y = 0: & & 8 & = 2 (x - 3)^2\\ \text{Divide both sides by} \ 2: & & 4 & = (x - 3)^2 \\ \text{Take the square root of both sides}: & & 4 & = x - 3 && \text{and} && -4 = x - 3\\ \text{Simplify}: & & & \underline{\underline{x = 7}} && \text{and} && \underline{\underline{x = -1}}\end{align*}

To find the y\begin{align*}y-\end{align*}intercept,

Set x=0:Simplify:y+8y+8=2(3)2=18y=10\begin{align*}\text{Set} \ x = 0: & & y + 8 & = 2(-3)^2\\ \text{Simplify}: & & y + 8 & = 18 \Rightarrow \underline{\underline{y = 10}}\end{align*}

To graph a parabola, we only need to know the following information:

• the vertex
• the x\begin{align*}x-\end{align*}intercepts
• the y\begin{align*}y-\end{align*}intercept
• whether the parabola turns up or down (remember that it turns up if a>0\begin{align*}a > 0\end{align*} and down if a<0\begin{align*}a < 0\end{align*})

#### Example B

Graph the parabola given by the function y+1=(x+3)2\begin{align*}y + 1 = (x +3)^2\end{align*}.

Solution

Rewrite:Vertex:y(1)=(x(3))2(3,1)vertex:(3,1)\begin{align*}& \text{Rewrite}: & & y - (-1) = (x - (-3))^2\\ & \text{Vertex}: & & \underline{(-3, -1)} && \text{vertex}:(-3, -1)\end{align*}

To find the x\begin{align*}x-\end{align*}intercepts,

Set y=0:Take the square root of both sides:Simplify:1=(x+3)21=x+3and1=x+3x=2and x=4xintercepts: (2,0) and (4,0)\begin{align*}&\text{Set} \ y = 0: & & 1 = ( x + 3)^2\\ &\text{Take the square root of both sides}: & & 1 = x + 3 \qquad \text{and} \qquad -1 = x + 3\\ &\text{Simplify}: && \underline{\underline{x = -2}} \qquad \quad \text{and} \qquad \quad \ \underline{\underline{x = -4}}\\ &&& x-\text{intercepts}: \ (-2, 0) \ \text{and} \ (-4, 0)\end{align*}

To find the y\begin{align*}y-\end{align*}intercept,

Set x=0:Simplify:y+1=(3)2y=8yintercept:(0,8)\begin{align*}& \text{Set} \ x = 0: & & y + 1 = (3)^2\\ & \text{Simplify:} & & \underline{\underline{y = 8}} && y-\text{intercept}: (0, 8)\end{align*}

And since a>0\begin{align*}a > 0\end{align*}, the parabola turns up.

Graph all the points and connect them with a smooth curve:

#### Example C

Graph the parabola given by the function y=12(x2)2\begin{align*}y = - \frac{1}{2} (x - 2)^2\end{align*}.

Solution:

RewriteVertex:y(0)=12(x2)2(2,0)vertex:(2,0)\begin{align*}& \text{Rewrite} & & y - (0) = - \frac{1} {2} (x - 2)^2\\ & \text{Vertex:} & & \underline{(2, 0)} && \text{vertex:} (2, 0)\end{align*}

To find the x\begin{align*}x-\end{align*}intercepts,

Set y=0:Multiply both sides by 2:Take the square root of both sides:Simplify:000=12(x2)2=(x2)2=x2x=2xintercept:(2,0)\begin{align*}\text{Set} \ y = 0: & & 0 & = - \frac{1} {2} (x - 2)^2 \\ \text{Multiply both sides by} \ -2: & & 0 & = (x - 2)^2 \\ \text{Take the square root of both sides}: & & 0 & = x - 2\\ \text{Simplify}: & & & \underline{\underline{x = 2}} && x-\text{intercept:} (2, 0)\end{align*}

Note: there is only one x\begin{align*}x-\end{align*}intercept, indicating that the vertex is located at this point, (2, 0).

To find the y\begin{align*}y-\end{align*}intercept,

Set x=0:Simplify:yy=12(2)2=12(4)y=2yintercept:(0,2)\begin{align*}\text{Set} \ x = 0: & & y & = -\frac{1} {2}(-2)^2 \\ \text{Simplify:} & & y & = - \frac{1} {2} (4) \Rightarrow \underline{\underline{y = -2}} && y- \text{intercept:}(0, -2)\end{align*}

Since a<0\begin{align*}a < 0\end{align*}, the parabola turns down.

Graph all the points and connect them with a smooth curve:

Watch this video for help with the Examples above.

### Vocabulary

• The vertex form of a quadratic function is

yk=a(xh)2\begin{align*}y - k = a(x - h)^2\end{align*}

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point (h,k)\begin{align*}(h, k)\end{align*}.

• To find the x\begin{align*}x-\end{align*}intercepts from the vertex form: just set y=0\begin{align*}y = 0\end{align*} and take the square root of both sides of the resulting equation.
• To find the y\begin{align*}y-\end{align*}intercept, set x=0\begin{align*}x = 0\end{align*} and simplify.

### Guided Practice

Graph the parabola given by the function y=4(x+2)21\begin{align*}y = 4(x +2)^2-1\end{align*}.

Solution:

RewriteSimplifyVertex:y(1)=4(x+2)2y+1=4(x+2)2(2,1)vertex:(2,1)\begin{align*}& \text{Rewrite} & & y - (-1) = 4(x +2)^2\\ & \text{Simplify} & & y +1 = 4(x +2)^2\\ & \text{Vertex:} & & \underline{(-2, -1)} && \text{vertex:} (-2, -1)\end{align*}

To find the \begin{align*}x-\end{align*}intercepts,

\begin{align*}\text{Set.} \ y = 0: & & 0 & = 4(x +2)^2-1 \\ \text{Subtract 1 from each side}: & & 1 & = 4(x +2)^2 \\ \text{Divide both sides by 4}: & & \frac{1}{4} & = (x +2)^2 \\ \text{Take the square root of both sides}: & & \frac{1}{2} & = \pm (x + 2)\\ \text{Separate}: & & & \frac{1}{2}=-(x+2) && \frac{1}{2}=x+2)\\ \text{Simplify}: & & & \underline{\underline{x = -2.5}} && \underline{\underline{x = -1.5}}\end{align*}

The \begin{align*}x-\end{align*}intercepts are \begin{align*}(-2.5, 0)\end{align*} and \begin{align*}(-1.5, 0)\end{align*}.

To find the \begin{align*}y-\end{align*}intercept,

\begin{align*}\text{Set} \ x = 0: & & y & = 4(0 +2)^2-1 \\ \text{Simplify:} & & y & = 15 \Rightarrow \underline{\underline{y = 15}} && y- \text{intercept:}(0, 15)\end{align*}

Since \begin{align*}a < 0\end{align*}, the parabola turns up.

Graph all the points and connect them with a smooth curve:

### Practice

Rewrite each quadratic function in vertex form.

1. \begin{align*} y= x^2 - 6x\end{align*}
2. \begin{align*}y + 1 = -2x^2 -x\end{align*}
3. \begin{align*}y = 9x^2 + 3x - 10\end{align*}
4. \begin{align*}y = -32x^2 + 60x + 10\end{align*}

For each parabola, find the vertex; the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts; and if it turns up or down. Then graph the parabola.

1. \begin{align*}y - 4 = x^2 + 8x\end{align*}
2. \begin{align*}y = -4x^2 + 20x - 24\end{align*}
3. \begin{align*}y = 3x^2 + 15x\end{align*}
4. \begin{align*}y + 6 = -x^2 + x\end{align*}
5. \begin{align*}x^2-10x+25=9\end{align*}
6. \begin{align*}x^2+18x+81=1\end{align*}
7. \begin{align*}4x^2-12x+9=16\end{align*}
8. \begin{align*}x^2+14x+49=3\end{align*}
9. \begin{align*}4x^2-20x+25=9\end{align*}
10. \begin{align*}x^2+8x+16=25\end{align*}

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### Vocabulary Language: English

TermDefinition
Intercept The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.
Parabola A parabola is the characteristic shape of a quadratic function graph, resembling a "U".
Vertex A vertex is a corner of a three-dimensional object. It is the point where three or more faces meet.

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