How do you write y=9x^2+3x-10y=9x2+3x−10 in vertex form?

1 Answer
Wataru
Nov 17, 2014

Recall

x^2+2ax+a^2=(x-a)^2x2+2ax+a2=(x−a)2

y=9x^2+3x-10y=9x2+3x−10

by factoring 9 out of the first two terms,

=9(x^2+1/3x)-10=9(x2+13x)−10

since 2a=1/3 => a=1/6 => a^2=1/362a=13⇒a=16⇒a2=136, by adding and subtracting 1/36136,

=9(x^2+1/3x+1/36-1/36)-10=9(x2+13x+136−136)−10

by keeping the first three term in the parentheses,

=9(x^2+1/3x+1/36)-9/36-10=9(x2+13x+136)−936−10

by completing the square,

=9(x+1/6)^2-41/4=9(x+16)2−414

I hope that this was helpful.

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