What is the vertex of $y= (x-3)^2-5x^2-2x-9$ ?

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Answer 1 · 2016-05-09T00:28:15

We will first have to expand and simplify the expression, before completing the square.

$y = x^2 - 6x + 9 - 5x^2 - 2x - 9$

$y = -4x^2 - 8x$

$y = -4(x^2 + 2x)$

$y = -4(x^2 + 2x + 1 - 1)$

$y = -4(x^2 + 2x + 1) -1(-4)$

$y = -4(x^2 + 2x + 1) + 4$

$y = -4(x + 1)^2 + 4$

As in any quadratic function of the form $y = a(x - p)^2 + q$ , the vertex is situated at the point $(p, q)$ .

Thus, the vertex is a $(-1, 4)$

Hopefully this helps!

What is the vertex of y= (x-3)^2-5x^2-2x-9 y= (x-3)^2-5x^2-2x-9?