# What is the vertex of  y= -8x^2+8x-(x+9)^2?

May 19, 2017

A sort of cheat method (not really)

color(blue)("Vertex"->(x,y)=(-5/9,-704/9)

#### Explanation:

Expanding the brackets we get:

$y = - 8 {x}^{2} + 8 x \text{ } - {x}^{2} - 18 x - 81$

$y = - 9 {x}^{2} - 10 x - 81 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

As the coefficient of ${x}^{2}$ is negative the graph is of form $\cap$
Thus the vertex is a maximum.

Consider the standardised form of $y = a {x}^{2} + b x + c$

Part of the process of completing the square is such that:

x_("vertex")=(-1/2)xxb/a" "=>" "(-1/2)xx((-10)/(-9)) = -5/9

Substitute for $x$ in $E q u a t i o n \left(1\right)$ giving:

${y}_{\text{vertex}} = - 9 {\left(- \frac{5}{9}\right)}^{2} - 10 \left(- \frac{5}{9}\right) - 81$

${y}_{\text{vertex}} = - 78 \frac{2}{9} \to - \frac{704}{9}$

color(blue)("Vertex"->(x,y)=(-5/9,-704/9)

Note that $- \frac{5}{9} \approx 0.55555 \ldots \to - 0.56$ to 2 decimal places 