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

1 Answer
evant939012
May 17, 2018

(-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).

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1947 views around the world
You can reuse this answer
Creative Commons License