# How do you write a quadratic equation with vertex at (-6,-2) and passes through point (-4,-14)?

Jul 29, 2017

Equation is $y = - 3 {\left(x + 6\right)}^{2} - 2$ or $x = \frac{1}{72} {\left(y + 2\right)}^{2} - 6$

#### Explanation:

A quadratic equation with a given vertex $\left(h , k\right)$ can be of two types

1. $y = a {\left(x - h\right)}^{2} + k$ or
2. $x = a {\left(y - k\right)}^{2} + h$

As such as vertex is $\left(- 6 , - 2\right)$

The equation is either $y = a {\left(x + 6\right)}^{2} - 2$ or $x = a {\left(y + 2\right)}^{2} - 6$

If it is $y = a {\left(x + 6\right)}^{2} - 2$ as the curve also passes through $\left(- 4 , - 14\right)$,

$- 14 = a {\left(- 4 + 6\right)}^{2} - 2$ or $4 a - 2 = - 14$ or $a = - 3$

and equation is $y = - 3 {\left(x + 6\right)}^{2} - 2$

If it is $x = a {\left(y + 2\right)}^{2} - 6$ as the curve also passes through $\left(- 4 , - 14\right)$,

$- 4 = a {\left(- 14 + 2\right)}^{2} - 6$ or $144 a - 6 = - 4$ or $a = \frac{1}{72}$

and equation is $x = \frac{1}{72} {\left(y + 2\right)}^{2} - 6$

graph{(y+3(x+6)^2+2)(72x-(y+2)^2+432)((x+6)^2+(y+2)^2-0.06)((x+4)^2+(y+14)^2-0.06)=0 [-19.67, 20.33, -19.28, 0.72]}