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)#.

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