Question

What is the vertex form of `y= y=x^2+5x-36`?

Answer 1

The vertex form `y--169/4=(x--5/2)^2`
with vertex at `(h, k)=(-5/2, -169/4)`

Explanation:

From the given equation `y=x^2+5x-36`

complete the square

`y=x^2+5x-36`
`y=x^2+5x+25/4-25/4-36`

We group the first three terms

`y=(x^2+5x+25/4)-25/4-36`

`y=(x+5/2)^2-25/4-144/4`

`y=(x+5/2)^2-169/4`

`y--169/4=(x--5/2)^2`

graph{y+169/4=(x--5/2)^2[-100, 100,-50,50]}

God bless...I hope the explanation is useful.

Answer 2

`y = (x + 5/2)^2 - 169/4`

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = -5/2`
y-coordinate of vertex:
`y(-5/2) = (25/4) - 25/2 - 36 = -25/4 - 36 = -169/4.`
`Vertex (-5/2, - 169/4)`
Vertex form: `y = (x + 5/2)^2 - 169/4`