How do you find the vertex of $y = 4x^2 + 8x + 7$ ?

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Answer 1 · 2016-03-28T07:52:14

$(h, k)=(-1, 3)$

From the given equation

$y=4x^2+8x+7$

There is a formula for finding the vertex $(h, k)$ for equation of the parabola of the form $y=ax^2+bx+c$

$h=-b/(2a)$ and $k=c-b^2/(4a)$

Solve for $h$ with $a=4$ and $b=8$ and $c=7$

$h=-b/(2a)=(-8)/(2*4)=-1$

Solve for $k$

$k=c-b^2/(4a)=7-8^2/(4*4)=7-64/16=7-4=3$

Our vertex is at $(-1, 3)$

See the graph of $y=4x^2+8x+7$ with vertex at $(-1, 3)$
graph{y=4x^2+8x+7[-20,20,-10,10]}

God bless....I hope the explanation is useful.

How do you find the vertex of y = 4x^2 + 8x + 7y = 4x^2 + 8x + 7?