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10.7: Vertex Form of a Quadratic Equation

Difficulty Level: At Grade Created by: CK-12
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What if you had a quadratic function like y2=x2+4x? 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.

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CK-12 Foundation: 1007S Graph Quadratic Functions 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

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point (h,k).

It is also simple to find the xintercepts from the vertex form: just set y=0 and take the square root of both sides of the resulting equation.

To find the yintercept, set x=0 and simplify.

Example A

Find the vertex, the xintercepts and the yintercept of the following parabolas:

a) y2=(x1)2

b) y+8=2(x3)2

Solution

a) y2=(x1)2

Vertex: (1, 2)

To find the xintercepts,

Set y=0:Take the square root of both sides:22=(x1)2=x1and2=x1

The solutions are not real so there are no xintercepts.

To find the yintercept,

Set x=0:Simplify:y2y2=(1)2=1y=3

b) y+8=2(x3)2

Rewrite:Vertex:y(8)=2(x3)2(3,8)

To find the xintercepts,

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

To find the yintercept,

Set x=0:Simplify:y+8y+8=2(3)2=18y=10

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

  • the vertex
  • the xintercepts
  • the yintercept
  • whether the parabola turns up or down (remember that it turns up if a>0 and down if a<0)

Example B

Graph the parabola given by the function y+1=(x+3)2.

Solution

Rewrite:Vertex:y(1)=(x(3))2(3,1)vertex:(3,1)

To find the xintercepts,

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)

To find the yintercept,

Set x=0:Simplify:y+1=(3)2y=8yintercept:(0,8)

And since a>0, 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.

Solution:

RewriteVertex:y(0)=12(x2)2(2,0)vertex:(2,0)

To find the xintercepts,

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)

Note: there is only one xintercept, indicating that the vertex is located at this point, (2, 0).

To find the yintercept,

Set x=0:Simplify:yy=12(2)2=12(4)y=2yintercept:(0,2)

Since a<0, the parabola turns down.

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

Watch this video for help with the Examples above.

CK-12 Foundation: 1007 Graph Quadratic Functions in Vertex Form

Vocabulary

  • The vertex form of a quadratic function is

yk=a(xh)2

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point (h,k).

  • To find the xintercepts from the vertex form: just set y=0 and take the square root of both sides of the resulting equation.
  • To find the yintercept, set x=0 and simplify.

Guided Practice

Graph the parabola given by the function y=4(x+2)21.

Solution:

RewriteSimplifyVertex:y(1)=4(x+2)2y+1=4(x+2)2(2,1)vertex:(2,1)

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*}

Vocabulary

Intercept

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

Parabola

A parabola is the characteristic shape of a quadratic function graph, resembling a "U".
Vertex

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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At Grade

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Date Created:

Oct 01, 2012

Last Modified:

Jan 31, 2016
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