How do you find the vertex of $y = x^2-8x+7$ ?

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Answer 1 · 2016-01-25T17:11:44

$V(4,-9)$

Factor

$y = x^2 -8x +7$ .

That's

$y = (x-4)^2 - 9$ .

Then, transpose $- 9$ to the other side of the equation, that would be

$y+9 = (x-4)^2$ .

Remember: $x$ is for h and $y$ is for k.

Vertex will be $V(4,-9)$ .

How do you find the vertex of y = x^2-8x+7y = x^2-8x+7?