Write a Quadratic Equation Given the Vertex and Another Point

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Question 1 of 2

1. Question

Find the equation of the graph below

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

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Vertex Form

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

where `(``h``,``k``)` is the vertex

Since the graph indicates the vertex, use the Vertex Form. Slot `(``h``,``k``)` and `(x,y)` into the Vertex Form to solve for `a`. Then, substitute `a`, `h` and `k` back to the main formula to form an equation.

First, label values from the graph

Note that the vertex is at `(-2,3)`

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

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

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

`a`	`=`	`-3/9`

`a`	`=`	`-1/3`	Simplify

Finally, substitute `a`, `h` and `k` into the Vertex Form

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

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

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

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

Question 2 of 2

2. Question