Question

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

Answer 1

A sort of cheat method (not really)

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

Explanation:

Expanding the brackets we get:

`y=-8x^2+8x" "-x^2-18x-81`

`y=-9x^2-10x-81" ".......................Equation(1)`

As the coefficient of `x^2` is negative the graph is of form `nn`
Thus the vertex is a maximum.

Consider the standardised form of `y=ax^2+bx+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 `Equation(1)` giving:

`y_("vertex")=-9(-5/9)^2-10(-5/9)-81`

`y_("vertex")=-78 2/9->-704/9`

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

Note that `-5/9~~0.55555... -> -0.56` to 2 decimal places