# How do you write the quadratic function in vertex form given vertex (-4,6) and point (-1,9)?

Apr 24, 2017

$3 y - 18 = {\left(x + 4\right)}^{2}$ or ${\left(y - 6\right)}^{2} = 3 \left(x + 4\right)$

#### Explanation:

A vertex form of equation is of the type

$\left(y - k\right) = a {\left(x - h\right)}^{2}$ or $\left(x - h\right) = a {\left(y - k\right)}^{2}$, where $\left(h , k\right)$ is the vertex.

As the vertex is $\left(- 4 , 6\right)$, it would be either

$\left(y - 6\right) = a {\left(x + 4\right)}^{2}$ or $\left(x + 4\right) = a {\left(y - 6\right)}^{2}$

Case 1

If it is $\left(y - 6\right) = a {\left(x + 4\right)}^{2}$, as it passes through $\left(- 1 , 9\right)$, we have

$\left(9 - 6\right) = a {\left(- 1 + 4\right)}^{2}$ i.e. $9 a = 3$ and $a = \frac{1}{3}$ and equation is

$\left(y - 6\right) = \frac{1}{3} {\left(x + 4\right)}^{2}$ i.e. $3 y - 18 = {\left(x + 4\right)}^{2}$

Case 2

If it is $\left(x + 4\right) = a {\left(y - 6\right)}^{2}$, as it passes through $\left(- 1 , 9\right)$, we have

$\left(- 1 + 4\right) = a {\left(9 - 6\right)}^{2}$ or $3 = 9 a$ and $a = \frac{1}{3}$ and equation is

$\left(x + 4\right) = \frac{1}{3} {\left(y - 6\right)}^{2}$ or ${\left(y - 6\right)}^{2} = 3 \left(x + 4\right)$

graph{(3y-18-(x+4)^2)((y-6)^2-3(x+4))=0 [-10.96, 9.04, 0.04, 10.04]}