# How do you write the equation of the parabola that has the same shape as y = x^2 but that has a vertex of (-3, 2)?

Jul 1, 2016

Consider vertex form, $y = a {\left(x - p\right)}^{2} + q$. The vertex is at $\left(p , q\right)$. Now consider the graph of $y = {x}^{2}$, as shown directly below. The vertex is at the origin, or in other words at $\left(0 , 0\right)$.

Since this graph is neither stretched nor compressed, the parameter a is $1$. Hence, the equation of this graph, in non simplified form is $y = 1 {\left(x - 0\right)}^{2} + 0$.

Knowing that parameter $a$ controls shape, and knowing that our new function has the same shape as $y = {x}^{2}$, we also know that $a = 1$ in our new equation. The vertex is at $\left(- 3 , 2\right)$, so replacing the $\left(0 , 0\right)$ inside the equation $y = 1 {\left(x - 0\right)}^{2} + 0$, we have $y = 1 {\left(x - \left(- 3\right)\right)}^{2} + 2$, or, better simplified, $y = {\left(x + 3\right)}^{2} + 2$.

Hence, the equation of the parabola with the same shape as $y = {x}^{2}$ and a vertex at $\left(- 3 , 2\right)$ is $y = {\left(x + 3\right)}^{2} + 2$.

Hopefully this helps!