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

$\left(h , k\right) = \left(- 1 , 3\right)$

#### Explanation:

From the given equation

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

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

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

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

$h = - \frac{b}{2 a} = \frac{- 8}{2 \cdot 4} = - 1$

Solve for $k$

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

Our vertex is at $\left(- 1 , 3\right)$

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

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