Question

Create the equation used to graph this graph:

Answer 1

`(y+4)^2=-4(x-2)` or
`x=-(1)/(4)(y+4)^2+2`

Explanation:

we know that

The general equation of a horizontal parabola open to the left is equal to

`(y-k)^2=-4p(x-h)`

where

(h,k) is the vertex of the parabola

p is the focal distance (distance from the vertex to the focus)

In this problem we have

The vertex is the point (2,-4)

so

`(h,k)=(2,-4)`

The Focus is the point F(1,-4)

so

the focal distance p=2-1=1 ---> distance from the vertex to the focus

substitute in the formula

`(y+4)^2=-4(1)(x-2)`

`(y+4)^2=-4(x-2)`

`x=-(1)/(4)(y+4)^2+2`