Question

What is the equation has a graph that is a parabola with a vertex at (-2, 0)?

Answer 1

A family of parabolas given by `(x+hy)^2+(2+c/2)x+by+c=0` . Upon setting h = 0, b =4 and c = 4, we get a member of the family as represented by `(x+2)^2=-4y`. The graph for this parabola is given.

Explanation:

The general equation of parabolas is

(x+hy)^2+ax+by+c=0. Note the perfect square for the 2nd degree

terms.

This passes through the vertex `(-2, 0)`. So,

`4-2a+c = 0 to a=2+c/2`

The required system ( family ) of parabolas is given by

`(x+hy)^2+(2+c/2)x+by+c=0`.

Let us get a member of the family.

Upon setting h = 0, b = c = 4, the equation becomes

`(x+2)^2=-4y`. The graph is inserted.

graph{-1/4 (x+2)^2 [-10, 10, -5, 5]}