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

May 17, 2018

$\left(- \frac{3}{8} , 129.125\right)$

#### Explanation:

Method A is completing the square.

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

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.

$f \left(x\right) = - 8 {x}^{2} - 6 x + 128$
$f ' \left(x\right) = - 16 x - 6$ This gives the slope of $f \left(x\right)$ at x.
$- 16 x - 6 = 0$ Find where the slope is zero, which is where the maximum is.
$x = - \frac{3}{8}$.

Substitute this back into the original equation to get 129.125, so the vertex is $\left(- \frac{3}{8} , 129.125\right)$.