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

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Answer 1 · 2016-06-23T15:36:33

Vertex $->(x,y)=(7/2, -45/2)$

Multiply out the bracket so that you combine terms as appropriate.

$y=x^2-6x+3" "-x-2$

$y=x^2-7x+1$

As the coefficient of $x^2$ is 1 we can apply directly

$x_("vertex")=(-1/2)xx(-7)$ where the -7 is from $-7x$

$x_("vertex")=+7/2$

Substitute in equation giving

$y_("vertex")=(7/2)^2-7(7/2)+1$

$y_("vertex")=-11 1/4 ->-45/4$

Tony B

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