What is the vertex of $y= x^2-x-16+(x-1)^2$ ?

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Answer 1 · 2017-12-13T20:23:39

First, expand the expression and combine like terms:

$x^2-x-16+(x-1)^2$

$\implies x^2-x-16+(x^2-2x+1)$

$\implies x^2+x^2-x-2x-16+1$

$\implies 2x^2-3x-15$

Now that's in the form $ax^2+bx+c$ , the vertex's $x$ -coordinate is $\frac{-b}{2a}$ .

$\implies \frac{3}{4}$

Plug that into the original equation to find the $y$ -coordinate:

$2x^2-3x-15$

$\implies 2(3/4)^2-3(3/4)-15$

$\implies 9/8-9/4-15/1$

$implies -16.125$

I'm in class rn and will finish this later. Sorry. :/

What is the vertex of y= x^2-x-16+(x-1)^2 y= x^2-x-16+(x-1)^2?