# How do I find the vertex of y=(x+7)^2?

The equation is in vertex form ($y = a {\left(x - h\right)}^{2} + k$) where the vertex is (h, k). Since the original form has you subtracting h, h more than likely has an opposite sign then what is in the equation. In the example, 7 is being added so therefore h = - 7. Since nothing is being added to the equation, k = 0. Therefore the vertex of $y = {\left(x + 7\right)}^{2}$ is $\left(- 7 , 0\right)$.
$y = - 2 {\left(x - 4\right)}^{2} - 3$ has h = 4 and k= -3, so the vertex is $\left(4 , - 3\right)$.
$y = 4 {\left(x + 1\right)}^{2} + 5$ has h = -1 and k = 5, so the vertex is $\left(- 1 , 5\right)$.