Graph Quadratic Functions by Completing the Square

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Question 1 of 4

Graph the function

y = x^{2} - 10 x + 15

$y=x^2-10x+15$

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Vertex Form

y = a {(x - h)}^{2} + k

$y=a(x-h)^2+k$

where

(h, k)

$(h,k)$ is the vertex

Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.

Plot the vertex and

x

$x$ intercepts, then connect the points to form a parabola

First, find the vertex and plot this on the graph

Start by transforming the function into vertex form

$y$ $y$	$=$ $=$	$x^{2} - 10 x + 15$ $x^2-10x+15$
$y$ $y$ $- 15$ $-15$	$=$ $=$	$x^{2} - 10 x + 15$ $x^2-10x+15$ $- 15$ $-15$	Subtract $15$ $15$ from both sides
$y - 15$ $y-15$	$=$ $=$	$x^{2} - 10 x$ $x^2-10x$

Take the coefficient of the

x

$x$ term, divide it by two and then square it.

$y - 15$ $y-15$	$=$ $=$	$x^{2}$ $x^2$ $- 10$ $-10$ $x$ $x$			Coefficient of the $x$ $x$ term

	$=$ $=$	$\frac{\color{#00880A}{-10}}{2}$			Divide it by $2$ $2$

		$(-5)^2$	$=$ $=$	$25$ $25$	Square

This number will make the right side a perfect square.

Add and subtract $25$ $25$ to the right side to keep the balance.

$y - 15$ $y-15$	$=$ $=$	$x^{2} - 10 x$ $x^2-10x$
$y - 15$ $y-15$	$=$ $=$	$x^{2} - 10 x +$ $x^2-10x+$ $25$ $25$ $-$ $-$ $25$ $25$	Add and subtract $25$ $25$

Now, transform the right side into a square of a binomial, then leave

y

$y$ on the left side.

$y - 15$ $y-15$	$=$ $=$	$(x - 5) (x - 5) - 25$ $(x-5)(x-5)-25$
$y - 15$ $y-15$	$=$ $=$	${(x - 5)}^{2} - 25$ $(x-5)^2-25$
$y - 15$ $y-15$ $+ 15$ $+15$	$=$ $=$	${(x - 5)}^{2} - 25$ $(x-5)^2-25$ $+ 15$ $+15$
$y$ $y$	$=$ $=$	${(x - 5)}^{2} - 10$ $(x-5)^2-10$

The function is now in vertex form

Compare the function to the general vertex form to get the vertex

$y$ $y$	$=$ $=$	$a {(x - h)}^{2} + k$ $a(x-h)^2+k$

$y$ $y$	$=$ $=$	${(x - 5)}^{2} - 10$ $(x-5)^2-10$

h

$h$

=

$=$

5

$5$

k

$k$

=

$=$

- 10

$-10$

This means that the vertex is at

(5, - 10)

$(5,-10)$

Next, find the

x

$x$ intercepts by substituting

y = 0

$y=0$ , then solving for

x

$x$

$y$ $y$	$=$ $=$	${(x - 5)}^{2} - 10$ $(x-5)^2-10$
$0$ $0$	$=$ $=$	${(x - 5)}^{2} - 10$ $(x-5)^2-10$	Substitute $y = 0$ $y=0$
$0$ $0$ $+ 10$ $+10$	$=$ $=$	${(x - 5)}^{2} - 10$ $(x-5)^2-10$ $+ 10$ $+10$	Add $10$ $10$ to both sides
$10$ $10$	$=$ $=$	${(x - 5)}^{2}$ $(x-5)^2$
$\sqrt{10}$ $sqrt10$	$=$ $=$	$\sqrt{{(x - 5)}^{2}}$ $sqrt((x-5)^2)$	Take the square root of both sides
$\pm \sqrt{10}$ $+-sqrt10$	$=$ $=$	$x - 5$ $x-5$
$x - 5$ $x-5$	$=$ $=$	$\pm \sqrt{10}$ $+-sqrt10$
$x - 5$ $x-5$ $+ 5$ $+5$	$=$ $=$	$\pm \sqrt{10}$ $+-sqrt10$ $+ 5$ $+5$	Add $5$ $5$ to both sides
$x$ $x$	$=$ $=$	$5 \pm \sqrt{10}$ $5+-sqrt10$
$x$ $x$	$=$ $=$	$8.162, 1.838$ $8.162, 1.838$

Mark these

2

$2$ points on the

x

$x$ axis

Finally, connect the points to form a parabola

Question 2 of 4

Graph the function

y = - 2 x^{2} - 8 x + 2

$y=-2x^2-8x+2$

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Vertex Form

y = a {(x - h)}^{2} + k

$y=a(x-h)^2+k$

where

(h, k)

$(h,k)$ is the vertex

Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.

Plot the vertex and

x

$x$ intercepts, then connect the points to form a parabola

First, find the vertex and plot this on the graph

Start by transforming the function into vertex form

$y$ $y$	$=$ $=$	$- 2 x^{2} - 8 x + 2$ $-2x^2-8x+2$
$y$ $y$	$=$ $=$	$- 2 (x^{2} + 4 x - 1)$ $-2(x^2+4x-1)$	Factor out $- 2$ $-2$

Take the coefficient of the

x

$x$ term, divide it by two and then square it.

$y$ $y$	$=$ $=$	$- 2 (x^{2} +$ $-2(x^2+$ $4$ $4$ $x - 1)$ $x-1)$			Coefficient of the $x$ $x$ term
	$=$ $=$	$\frac{\color{#00880A}{4}}{2}$			Divide it by $2$ $2$

		$(2)^2$	$=$ $=$	$4$ $4$	Square

This number will make the right side a perfect square.

Add and subtract $4$ $4$ to the grouping of

x

$x$ to keep the balance.

$y$ $y$	$=$ $=$	$- 2 (x^{2} + 4 x - 1)$ $-2(x^2+4x-1)$
$y$ $y$	$=$ $=$	$- 2 (x^{2} + 4 x +$ $-2(x^2+4x+$ $4$ $4$ $-$ $-$ $4$ $4$ $- 1)$ $-1)$	Add and subtract $4$ $4$
$y$ $y$	$=$ $=$	$- 2 ({(x + 2)}^{2} - 4 - 1)$ $-2((x+2)^2-4-1)$
$y$ $y$	$=$ $=$	$- 2 ({(x + 2)}^{2} - 5)$ $-2((x+2)^2-5)$
$y$ $y$	$=$ $=$	$- 2 {(x + 2)}^{2} + 10$ $-2(x+2)^2+10$

This is now in vertex form

Compare the function to the general vertex form to get the vertex

$y$ $y$	$=$ $=$	$a {(x - h)}^{2} + k$ $a(x-h)^2+k$
$y$ $y$	$=$ $=$	$- 2 {(x + 2)}^{2} + 10$ $-2(x+2)^2+10$

h

$h$

=

$=$

- 2

$-2$

k

$k$

=

$=$

10

$10$

This means that the vertex is at

(- 2, 10)

$(-2,10)$

Next, find the

x

$x$ intercepts by substituting

y = 0

$y=0$ , then solving for

x

$x$

$y$ $y$	$=$ $=$	$- 2 {(x + 2)}^{2} + 10$ $-2(x+2)^2+10$
$0$ $0$	$=$ $=$	$- 2 {(x + 2)}^{2} + 10$ $-2(x+2)^2+10$	Substitute $y = 0$ $y=0$
$0$ $0$ $- 10$ $-10$	$=$ $=$	$- 2 {(x + 2)}^{2} + 10$ $-2(x+2)^2+10$ $- 10$ $-10$	Subtract $10$ $10$ from both sides
$- 10$ $-10$	$=$ $=$	$- 2 {(x + 2)}^{2}$ $-2(x+2)^2$
$- 10$ $-10$ $\div (- 2)$ $-:(-2)$	$=$ $=$	$- 2 {(x + 2)}^{2}$ $-2(x+2)^2$ $\div (- 2)$ $-:(-2)$	Divide both sides by $- 2$ $-2$
$5$ $5$	$=$ $=$	${(x + 2)}^{2}$ $(x+2)^2$
${(x + 2)}^{2}$ $(x+2)^2$	$=$ $=$	$5$ $5$
$\sqrt{{(x + 2)}^{2}}$ $sqrt((x+2)^2)$	$=$ $=$	$\sqrt{5}$ $sqrt5$	Take the square root of both sides
$x + 2$ $x+2$	$=$ $=$	$\pm \sqrt{5}$ $+-sqrt5$
$x + 2$ $x+2$ $- 2$ $-2$	$=$ $=$	$\pm \sqrt{5}$ $+-sqrt5$ $- 2$ $-2$	Subtract $2$ $2$ from both sides
$x$ $x$	$=$ $=$	$- 2 \pm \sqrt{5}$ $-2+-sqrt5$

Mark these

2

$2$ points on the

x

$x$ axis

Finally, connect the points to form a parabola

Question 3 of 4

By completing the square, which graph is correct for the equation:

y = x^{2} - 8 x + 24

$y=x^2-8x+24$ .

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Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.

Perform the process of completing the square on the given quadratic to convert into vertex form.

$y$ $y$	$=$ $=$	$x^{2} - 8 x + 24$ $x^2-8x+24$
	$=$ $=$	$(x^{2} - 8 x$ $(x^2-8x$ $+ {(- \frac{8}{2})}^{2}$ $+ (-8/2)^2$ $)$ $)$ $- {(- \frac{8}{2})}^{2}$ $-(-8/2)^2$ $+ 24$ $+ 24$	Complete the square

	$=$ $=$	$(x^{2} - 8 x$ $(x^2-8x$ $+ 16$ $+16$ $)$ $)$ $- 16$ $-16$ $+ 24$ $+ 24$	Simplify

	$=$ $=$	$(x^{2} - 8 x$ $(x^2-8x$ $+ 16$ $+16$ $) + 8$ $)+8$
$y$ $y$	$=$ $=$	${(x - 4)}^{2} + 8$ $(x-4)^2+8$	Rewrite as a square of a binomial

Identify the vertex of the graph from the given formula.

$y$ $y$	$=$ $=$	$a(x-\color{blue}{h})^2+\color{blue}{k}$
$y$ $y$	$=$ $=$	${(x - 4)}^{2} + 8$ $(x-4)^2+8$	Given equation
$y$ $y$	$=$ $=$	$a(x-\color{blue}{4})^2+\color{blue}{8}$	Extract values of $h$ $h$ and $k$ $k$
Vertex	$=$ $=$	$(\color{blue}{h},\color{blue}{k})$
Vertex	$=$ $=$	$(\color{blue}{4},\color{blue}{8})$

Mark the vertex on the graph.

Next, solve for the

y

$y$ -intercept by substituting $x = 0$ $x=0$ .

$y$ $y$	$=$ $=$	${(x - 4)}^{2} + 8$ $(x-4)^2+8$
$y$ $y$	$=$ $=$	${(0 - 4)}^{2} + 8$ $(0-4)^2+8$	Substitute $x = 0$ $x=0$
$y$ $y$	$=$ $=$	$16 + 8$ $16+8$
$y$ $y$	$=$ $=$	$24$ $24$

Mark the

y

$y$ -intercept on the graph.

Draw a parabola using the points.

Question 4 of 4

By completing the square, which graph is correct for the equation:

y = x^{2} - 8 x + 18

$y=x^2-8x+18$ .

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Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.

Perform the process of completing the square on the given quadratic to convert into vertex form.

$y$ $y$	$=$ $=$	$x^{2} - 8 x + 18$ $x^2-8x+18$
	$=$ $=$	$(x^{2} - 8 x$ $(x^2-8x$ $+ {(- \frac{8}{2})}^{2}$ $+ (-8/2)^2$ $)$ $)$ $- {(- \frac{8}{2})}^{2}$ $-(-8/2)^2$ $+ 18$ $+ 18$	Complete the square

	$=$ $=$	$(x^{2} - 8 x$ $(x^2-8x$ $+ 16$ $+16$ $)$ $)$ $- 16$ $-16$ $+ 18$ $+ 18$	Simplify

	$=$ $=$	$(x^{2} - 8 x$ $(x^2-8x$ $+ 16$ $+16$ $) + 2$ $)+2$
$y$ $y$	$=$ $=$	${(x - 4)}^{2} + 2$ $(x-4)^2+2$	Rewrite as a square of a binomial

Identify the vertex of the graph from the given formula.

$y$ $y$	$=$ $=$	$a(x-\color{blue}{h})^2+\color{blue}{k}$
$y$ $y$	$=$ $=$	${(x - 4)}^{2} + 2$ $(x-4)^2+2$	Given equation
$y$ $y$	$=$ $=$	$a(x-\color{blue}{4})^2+\color{blue}{2}$	Extract values of $h$ $h$ and $k$ $k$
Vertex	$=$ $=$	$(\color{blue}{h},\color{blue}{k})$
Vertex	$=$ $=$	$(\color{blue}{4},\color{blue}{2})$

Mark the vertex on the graph.

Next, solve for the

y

$y$ -intercept by substituting $x = 0$ $x=0$ .

$y$ $y$	$=$ $=$	${(x - 4)}^{2} + 2$ $(x-4)^2+2$
$y$ $y$	$=$ $=$	${(0 - 4)}^{2} + 2$ $(0-4)^2+2$	Substitute $x = 0$ $x=0$
$y$ $y$	$=$ $=$	$16 + 2$ $16+2$
$y$ $y$	$=$ $=$	$18$ $18$

Mark the

y

$y$ -intercept on the graph.

Draw a parabola using the points.

Graph Quadratic Functions by Completing the Square

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Quiz summary

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1. Question

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Excellent!

Vertex Form

2. Question

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Exceptional!

Vertex Form

3. Question

Hint

Well Done!

4. Question

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Nice Job!

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