Question

How do you write `f(x)= -2x^2 + 16x +4` in vertex form?

Answer 1

`f(x)=-2(x-4)^2+36` with vertex at `(4,36)`

Explanation:

General vertex form is
`color(white)("XXX")f(x)=color(green)(m)(x-color(red)(a))^2+color(blue)(b)` with the vertex at `(color(red)(a),color(blue)(b))`

Given
`color(white)("XXX")f(x)=-2x^2+16x+4`

Extract the `color(green)m` component:
`color(white)("XXX")f(x)=color(green)(-2)(x^2-8x)+4`

Complete the square:
`color(white)("XXX")f(x)=color(green)(-2)(x^2-8x+color(cyan)(16))+4-color(cyan)((-2*)(16))`

Re-write as squared binomial and simplify to get vertex form
`color(white)("XXX")f(x)=color(green)(-2)(x-color(red)(4))^2+color(blue)(36)`
graph{-2x^2+16x+4 [-5.92, 26.13, 22.5, 38.54]}

Answer 2

The vertex form is `(x-4)^2=-1/2*(y-36)`

Explanation:

We start from the given `f(x)=-2x^2+16x+4`

Let `y=-2x^2+16x+4`

Start by factoring out the -2 from the first two terms

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

We now use the -8. Divide this number by 2 then the result be squared so that we will have `(-8/2)^2=+16`

This 16 will be added and subtracted inside the grouping symbol.

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

`y=-2(x^2-8x+16-16)+4`

We now have a PFT-Perfect Square Trinomial `(x^2-8x+16)=(x-4)^2`

So that we have

`y=-2(x^2-8x+16-16)+4`

`y=-2((x-4)^2-16)+4`

Put the -2 back

`y=-2(x-4)^2+32+4`

Simplify

`y=-2(x-4)^2+36`
transpose the 36 to the left of the equation

`y-36=-2(x-4)^2`

divide by -2

`(x-4)^2=-1/2*(y-36)`

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

Answer 3

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

Explanation:

There is another way of finding the vertex form.
x-coordinate of vertex:
`x = -b/(2a) = -16/-4 = 4`
y-coordinate of vertex:
y(4) = -32 + 64 + 4 = 36
Vertex form:
`y = a(x - 4)^2 + 36 = -2(x - 4)^2 + 36`