Question

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

Answer 1

Equation is `y=-3(x+6)^2-2` or `x=1/72(y+2)^2-6`

Explanation:

A quadratic equation with a given vertex `(h,k)` can be of two types

1. `y=a(x-h)^2+k` or
2. `x=a(y-k)^2+h`

As such as vertex is `(-6,-2)`

The equation is either `y=a(x+6)^2-2` or `x=a(y+2)^2-6`

If it is `y=a(x+6)^2-2` as the curve also passes through `(-4,-14)`,

`-14=a(-4+6)^2-2` or `4a-2=-14` or `a=-3`

and equation is `y=-3(x+6)^2-2`

If it is `x=a(y+2)^2-6` as the curve also passes through `(-4,-14)`,

`-4=a(-14+2)^2-6` or `144a-6=-4` or `a=1/72`

and equation is `x=1/72(y+2)^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]}