What is the vertex of $y=8(3x+7)^2+5$ ?

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Answer 1 · 2017-10-20T17:45:48

$(-7/3, 5)=(-2.bar(3), 5)$

First get this into vertex form:

$y=a(b(x-h))^2+k$ where $(h,k)$ is the vertex

by factoring out the $3$ in the parentheses:

$y=8(3(x+7/3))^2+5$

Then factor out a negative $1$ :

$y=8(3(x-1(-7/3)))^2+5$

So it's now in vertex form:

$y=8(3(x-(-7/3)))^2+5$ where $h=-7/3$ and $k=5$

So our vertex is $(-7/3, 5)=(-2.bar(3), 5)$

What is the vertex of y=8(3x+7)^2+5y=8(3x+7)^2+5?