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!