What is the vertex of $y= (x-3)^2-5x^2-x-1$ ?

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Answer 1 · 2017-11-10T20:19:29

The vertex is at $(-7/8, 177/16)$

The equation given is a quadratic $y = ax^2 + bx +c$

The vertex is at $(h,k)$ where $h = -b/(2a)$

First expand the equation
$y = x^2 - 6x + 9 -5x^2 -x -1$

Simplify
$y = -4x^2 -7x +8$

the $x$ value of the vertex is $7/-8$ or $-7/8$

plug the value for h back into the equation to get k

$y = -4*-7/8*-7/8 -7*-7/8 +8$ = $177/16$

The vertex is at $(-7/8, 177/16)$

What is the vertex of y= (x-3)^2-5x^2-x-1 y= (x-3)^2-5x^2-x-1?