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

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Answer 1 · 2017-05-19T11:31:56

A sort of cheat method (not really)

$Vertex \to (x, y) = (- \frac{5}{9}, - \frac{704}{9})$ $color(blue)("Vertex"->(x,y)=(-5/9,-704/9)$

Expanding the brackets we get:

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

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

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
Tony B Tony B

What is the vertex of y= -8x^2+8x-(x+9)^2y=−8x2+8x−(x+9)2 y= -8x^2+8x-(x+9)^2?