Question

Rewrite y=x^2-8x+15 in vertex form. Then state whether the vertex is a maximum or a minimum and give it´s coordinates.

Answer 1

Answer:

$\large \boxed{\sf \ \ y=(x-4)^2-1 \ \ }$

Step-by-step explanation:

Hello,

$y = x^2-8x+15\\ \\\text{*** we need to complete the square ***} \\ \\y=(x-4)^2-4^2+15=(x-4)^2-16+15=(x-4)^2-1$

And then the vertex is (4, -1) and as a > 0 the vertex is a minimum.

Hope this helps.

Do not hesitate if you need further explanation.

Thank you