Write a Quadratic Equation Given the Vertex and Another Point

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Question 1 of 2

Find the equation of the graph below

1. $y = - \frac{1}{3} {(x + 2)}^{2} + 3$ $y=-1/3 (x+2)^2+3$
2. $y = - \frac{1}{3} {(x + 2)}^{2} - 3$ $y=-1/3 (x+2)^2-3$
3. $y = \frac{1}{3} {(x - 2)}^{2} + 3$ $y=1/3 (x-2)^2+3$
4. $y = \frac{1}{3} {(x + 2)}^{2} + 3$ $y=1/3 (x+2)^2+3$

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Vertex Form

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

$y=a{(x-\color{#007DDC}{h})}^2+\color{#e65021}{k}$

where

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ is the vertex

Since the graph indicates the vertex, use the Vertex Form. Slot

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ and

(x, y)

$(x,y)$ into the Vertex Form to solve for

a

$a$ . Then, substitute

a

$a$ , $h$ $h$ and $k$ $k$ back to the main formula to form an equation.

First, label values from the graph

Note that the vertex is at

(- 2, 3)

$(-2,3)$

$h$ $h$	$=$ $=$	$- 2$ $-2$	from vertex
$k$ $k$	$=$ $=$	$3$ $3$	from vertex
$x$ $x$	$=$ $=$	$- 5$ $-5$	$x$ $x$ intercept
$y$ $y$	$=$ $=$	$0$ $0$	value of $y$ $y$ at $x$ $x$ intercept

Now, slot these values into the Vertex Form and solve for

a

$a$

$y$ $y$	$=$ $=$	$a{(\color{#00880A}{x}-\color{#007DDC}{h})}^2+\color{#e65021}{k}$	Vertex Form
$0$ $0$	$=$ $=$	$a{(\color{#00880A}{-5}-(\color{#007DDC}{-2}))}^2+\color{#e65021}{3}$	Substitute values
$0$ $0$	$=$ $=$	$a {(- 3)}^{2} + 3$ $a(-3)^2+3$
$0$ $0$	$=$ $=$	$9 a + 3$ $9a+3$
$9 a + 3$ $9a+3$	$=$ $=$	$0$ $0$
$9 a + 3$ $9a+3$ $- 3$ $-3$	$=$ $=$	$0$ $0$ $- 3$ $-3$	Subtract $3$ $3$ from both sides
$9 a$ $9a$ $\div 9$ $divide9$	$=$ $=$	$- 3$ $-3$ $\div 9$ $divide9$	Divide both sides by $9$ $9$

$a$ $a$	$=$ $=$	$- \frac{3}{9}$ $-3/9$

$a$ $a$	$=$ $=$	$- \frac{1}{3}$ $-1/3$	Simplify

Finally, substitute

a

$a$ ,

h

$h$ and

k

$k$ into the Vertex Form

$y$ $y$	$=$ $=$	$a{(x-\color{#007DDC}{h})}^2+\color{#e65021}{k}$	Vertex Form

$y$ $y$	$=$ $=$	$-\frac{1}{3}{(x-(\color{#007DDC}{-2}))}^2+\color{#e65021}{3}$	Substitute values

$y$ $y$	$=$ $=$	$- \frac{1}{3} {(x + 2)}^{2} + 3$ $-1/3 (x+2)^2+3$

y = - \frac{1}{3} {(x + 2)}^{2} + 3

$y=-1/3 (x+2)^2+3$

Question 2 of 2

Find the equation of the graph below

1. $y = - {(x - 2)}^{2} + 3$ $y=-(x-2)^2+3$
2. $y = {(x - 3)}^{2} - 2$ $y=(x-3)^2-2$
3. $y = - {(x - 3)}^{2} + 2$ $y=-(x-3)^2+2$
4. $y = {(x + 2)}^{2} + 3$ $y=(x+2)^2+3$

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Vertex Form

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

$y=a{(x-\color{#007DDC}{h})}^2+\color{#e65021}{k}$

where

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ is the vertex

Since the graph indicates the vertex, use the Vertex Form. Slot

(

$($ $h$ $h$

,

$,$ $k$ $k$

)

$)$ and

(x, y)

$(x,y)$ into the Vertex Form to solve for

a

$a$ . Then, substitute

a

$a$ , $h$ $h$ and $k$ $k$ back to the main formula to form an equation.

First, label values from the graph

$h$ $h$	$=$ $=$	$3$ $3$	from vertex
$k$ $k$	$=$ $=$	$2$ $2$	from vertex
$x$ $x$	$=$ $=$	$4$ $4$	from given point
$y$ $y$	$=$ $=$	$1$ $1$	from given point

Now, slot these values into the Vertex Form and solve for

a

$a$

$y$ $y$	$=$ $=$	$a{(\color{#00880A}{x}-\color{#007DDC}{h})}^2+\color{#e65021}{k}$	Vertex Form
$1$ $1$	$=$ $=$	$a{(\color{#00880A}{4}-\color{#007DDC}{3})}^2+\color{#e65021}{2}$	Substitute values
$1$ $1$	$=$ $=$	$a {(1)}^{2} + 2$ $a(1)^2+2$
$1$ $1$	$=$ $=$	$a + 2$ $a+2$
$a + 2$ $a+2$	$=$ $=$	$1$ $1$
$a + 2$ $a+2$ $- 2$ $-2$	$=$ $=$	$1$ $1$ $- 2$ $-2$	Subtract $2$ $2$ from both sides
$a$ $a$	$=$ $=$	$- 1$ $-1$

Finally, substitute

a

$a$ ,

h

$h$ and

k

$k$ into the Vertex Form

$y$ $y$	$=$ $=$	$a{(x-\color{#007DDC}{h})}^2+\color{#e65021}{k}$	Vertex Form
$y$ $y$	$=$ $=$	$-1{(x-\color{#007DDC}{3})}^2+\color{#e65021}{2}$	Substitute values
$y$ $y$	$=$ $=$	$- {(x - 3)}^{2} + 2$ $-(x-3)^2+2$

y = - {(x - 3)}^{2} + 2

$y=-(x-3)^2+2$

Write a Quadratic Equation Given the Vertex and Another Point

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Quiz summary

Information

1. Question

Hint

Correct!

Vertex Form

2. Question

Hint

Excellent!

Vertex Form

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