Question

What is the vertex of `y= -8x^2 − 6x + 128`?

Answer 1

`(-3/8, 129.125)`

Explanation:

There are actually 2 methods of going about this.

Method A is completing the square.

To do this, the function needs to be in the form `y=a(x-h)^2+k`.
First, separate the constant from the first two terms:
`-8x^2-6x` `+128`
Then factor out -8:
`-8(x^2+6/8x)+128`
`6/8` can be reduced to `3/4`.
Next, divide the `3/4` by 2 and square it:
`-8(x^2+3/4x+9/64)`
Make sure to SUBTRACT `9/64 * -8` so that the equation remains the same.
`-8(x^2+3/4x+9/64)+128-(-9/8)`
Simplify to get:
`-8(x+3/8)^2+129.125`

Method 2: Calculus

There is a method that is sometimes easier or harder. It involves taking the derivative of the equation, setting it equal to 0, and substituting that solution back into the original equation.

**If you don't understand, don't worry. This method is harder for this specific question.

`f(x)=-8x^2-6x+128`
`f'(x)=-16x-6` This gives the slope of `f(x)` at x.
`-16x-6=0` Find where the slope is zero, which is where the maximum is.
`x=-3/8`.

Substitute this back into the original equation to get 129.125, so the vertex is `(-3/8, 129.125)`.