How do you write $y=9x^2+3x-10$ in vertex form?

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Answer 1 · 2014-11-17T23:27:00

Recall

$x^2+2ax+a^2=(x-a)^2$

$y=9x^2+3x-10$

by factoring 9 out of the first two terms,

$=9(x^2+1/3x)-10$

since $2a=1/3 => a=1/6 => a^2=1/36$ , by adding and subtracting $1/36$ ,

$=9(x^2+1/3x+1/36-1/36)-10$

by keeping the first three term in the parentheses,

$=9(x^2+1/3x+1/36)-9/36-10$

$=9(x+1/6)^2-41/4$

I hope that this was helpful.

How do you write y=9x^2+3x-10y=9x^2+3x-10 in vertex form?