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

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Answer 1 · 2016-03-28T07:52:14

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

From the given equation

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

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

$h = - \frac{b}{2 a}$ $h=-b/(2a)$ and $k = c - \frac{b^{2}}{4 a}$ $k=c-b^2/(4a)$

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

$h = - \frac{b}{2 a} = \frac{- 8}{2 \cdot 4} = - 1$ $h=-b/(2a)=(-8)/(2*4)=-1$

Solve for $k$ $k$

$k = c - \frac{b^{2}}{4 a} = 7 - \frac{8^{2}}{4 \cdot 4} = 7 - \frac{64}{16} = 7 - 4 = 3$ $k=c-b^2/(4a)=7-8^2/(4*4)=7-64/16=7-4=3$

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

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

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

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