3 years ago

Enter the values of h and I so that y=x2+6x+10 is in vertex form Y=(x+)^2+_

1 answer:

3 0

$Use\ (a+b)^2=a^2+2ab+b^2\qquad(*)\\\\y=x^2+6x+10=x^2+2(x)(3)+10=\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2+10\\\\=(x+3)^2-9+10=(x+3)^2+1\\\\Answer:\ \boxed{y=(x+3)^2+1}$

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