Question

What is the vertex of `y= 4(x+2)^2-2x^2-3x-1`?

Answer 1

Vertex : `(-13/4 , -49/8)`

Vertex form: `y = 2(x+13/4)^2 -49/8`

Explanation:

Step 1: Expand/multiply the function so it can be int eh standard form of `y = ax^2 + bc+c`

Given
`y = 4(x+2)^2 -2x -3x -1`

`= 4(x+2)(x+2)-2x^2 -3x-1`

`= 4(x^2 +2x+2x+4) -2x^2 -3x-1`

`=4(x^2 +4x+4) = 2x^2 -3x -1`

`=4x^2 +16 x +16 -2x^2 -3x -1`

`=2x^2 + 13x+15`

`a = 2, " " " b= 13, " " " c= 15`

The formula for vertex is `(-b/(2a), f(-b/(2a)))`

`x_(vertex) = -b/(2a) = h`

`x_(vertex) = (-13)/(2*2) = -13/4`

`y_(vertex) = f(-b/(2a)) = k`

`f(-13/4) = 2(-13/4)^2 +13(-13/4)+15`

`= 2(169/16)-169/4 +15 = -49/8`

Vertex form: `y = a(x-h)^2 + k`

`y = 2(x+13/4)^2 -49/8`