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

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

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Answer 1 · 2014-11-17T23:27:00

Recall

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

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

by factoring 9 out of the first two terms,

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

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

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

by keeping the first three term in the parentheses,

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

by completing the square,

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

I hope that this was helpful.

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

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