SOLUTION: Okay well I have broken down the equation of
Y=-5x^2+30x-36
Down to
Y=-5(x^2-6x)-36
I'm trying to break it down to standard form.
Then it's telling me to recall that be cons
Algebra ->
Quadratic-relations-and-conic-sections
-> SOLUTION: Okay well I have broken down the equation of
Y=-5x^2+30x-36
Down to
Y=-5(x^2-6x)-36
I'm trying to break it down to standard form.
Then it's telling me to recall that be cons
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Question 869742: Okay well I have broken down the equation of
Y=-5x^2+30x-36
Down to
Y=-5(x^2-6x)-36
I'm trying to break it down to standard form.
Then it's telling me to recall that be constant term in a perfect square is the square of half of the coefficient of the linear term therefore 9 is added inside the parentheses to form a perfect square
Leaving x^2-6x+9
My question is where exactly did the 9 come from? Found 2 solutions by mananth, josgarithmetic:Answer by mananth(16946) (Show Source):
Recall (a-b)^2=a^2-2ab+b^2
we have a^2-2ab
which is
x^2-(2)(3)x
adding 9 will make it a perfect square
x^2-(2)(3)(3)+9
we have added 9 so we have to subtract 9
x^2-6x+9-9
Finally we have
y= -5(x^2-6x+9-9) -36
y=-5((x-3)^2-9)-6^2
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