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

Nov 17, 2014

Recall

${x}^{2} + 2 a x + {a}^{2} = {\left(x - a\right)}^{2}$

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

by factoring 9 out of the first two terms,

$= 9 \left({x}^{2} + \frac{1}{3} x\right) - 10$

since $2 a = \frac{1}{3} \implies a = \frac{1}{6} \implies {a}^{2} = \frac{1}{36}$, by adding and subtracting $\frac{1}{36}$,

$= 9 \left({x}^{2} + \frac{1}{3} x + \frac{1}{36} - \frac{1}{36}\right) - 10$

by keeping the first three term in the parentheses,

$= 9 \left({x}^{2} + \frac{1}{3} x + \frac{1}{36}\right) - \frac{9}{36} - 10$

$= 9 {\left(x + \frac{1}{6}\right)}^{2} - \frac{41}{4}$

I hope that this was helpful.