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If we express $3x^2 - 6x - 2$ in the form $a(x - h)^2 + k$, then what is $a + h + k$?

 Feb 4, 2021

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 #1
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Let's use some clever algebraic manipulation to change this quadratic from standard from to vertex form.

 

\(f(x) = 3x^2 - 6x - 2\\ f(x) = 3(x^2 - 2x) - 2\\ f(x) = 3(x^2 - 2x + 1) - 2 - 3\\ f(x) = 3(x-1)^2 - 5\\ a = 3\text{ and } h = 1 \text{ and } k = -5\\ a+h+k = 3 + 1 - 5\\ a+h+k = -1\)

 Feb 4, 2021
 #1
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0
Best Answer

Let's use some clever algebraic manipulation to change this quadratic from standard from to vertex form.

 

\(f(x) = 3x^2 - 6x - 2\\ f(x) = 3(x^2 - 2x) - 2\\ f(x) = 3(x^2 - 2x + 1) - 2 - 3\\ f(x) = 3(x-1)^2 - 5\\ a = 3\text{ and } h = 1 \text{ and } k = -5\\ a+h+k = 3 + 1 - 5\\ a+h+k = -1\)

Guest Feb 4, 2021

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