How do I write a quadratic equation given the vertex and an intercept?

Question

Some examples you are free to use are:

1. Vertex: (2.3) y-intercept: 7
2. Vertex: (-3,2) y-intercept:-4

Answer 1

(x-2)^2 =2p(y-3)
So 4 = 2p(7-3) --> 8p=4 and p = 1/2
So equation is (x-2)^2 = .5(y-3)

Note 2p is the distance between the focus and the directrix with the vertex at the midpoint.

Answer 2

The standard equation of parabola

y = a(x-h)^2 + k

where h and k are vertex coordinates

h = 2, k= 3

substituting

y = a(x -2)^2 + 3

y intercept is obtained when x = 0

y = a(-2)^2 + 3

y = 4a + 3

4a+ 3 = 7

4a = 7-3 = 4

a = 4/4 = 1

so the equation is

y = (x - 2)^2 + 3

expanding

y = x^2 -4x + 4 + 3

y = x^2 - 4x + 7

2)

y = a(x+3)^2 +2


when x = 0

-4 = 9a + 2

9a = -6

a = -6/9 = -2/3

y = -2/3(x+3)^2 + 2

y = -2/3(x^2 + 6x + 9) +2

3y = -2x^2 - 12x - 18 + 6

3y = -2x^2 - 12x - 12