Question

Create the equation used to graph this graph:

Answer 1

Answer:

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

Step-by-step 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

$(y+4)^2=-4(1)(x-2)$  

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

$x=-\frac{1}{4}(y+4)^2+2$