Vertex form of a Parabola?

Question

Find the equaion of the parabola given the vertex V and another point P.

10. V (3,1) P (-1, 3)

Answer 1

If you know the vertex then you know

a (x-xv)^2=y-yv , use P to find a, the shape

at P:
a(xp-3)^2 =yp-1
a(-1-3)^2 = 3-1
16a = 2
a=1/8

So, the parabola is

y = 1/8*(x-3)^2 + 1

Check:
at V y= 1 ok
at P y = 1/8(-4)^2 + 1 = 16/8+1 = 3 ok

Answer 2

You need more information. There are 2 parabolas that can fit that, one that opens upward, and one that opens to the left.

I'm going to assume that you want the one upwards.

The general form of a parabola opening up or down is:

(x - h)² = a(y - k)

(a is sometimes written as 1/4p, but in this case a is fine.)

You're given the vertex (h,k) = (3,1) and the point (x,y) = (-1, 3). All you have to do is solve for a, and you've got the equation:

(-1 - 3)² = a(3 - 1)

Then just plug in that value for a in the original equation, solve it for y, and you'll have your answer.