find the equation?

Question

a parabola has intercepts at x=-8, x=4 and y=-8. What is the equation of the parabola?

how would we find the equation?? i dont want the asnwer, i just wnat to know how to find it.

Answer 1

First figure the x coordinate of the vertex. It will be halfway between the two x coords (-8 + 4) / 2 = -2

In vertex notation you have:
y - k = a(x - h)^2.

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

Now you are almost there. You need another pair of points, so try x = 0, y = -8 (the y-intercept).

-8 - k = a(0 + 2)^2
-8 - k = 4a
k = 8 - 4a

Then try an x-intercept (4, 0).

0 - k = a(4 + 2)^2
-k = 36a
k = -36a

Now equate k to solve for a:
-36a = -4a + 8
8 = 32a
a = 1/4

Now you are to:
y - k = 1/4(x + 2)^2

And finally you can solve for k:
k = -36a
k = -36(1/4)
k = -9

Here's the answer in vertex form:
y + 9 = 1/4(x + 2)^2

Or you can expand this out:
y = 1/4(x + 2)^2 - 9
y = 1/4x^2 + x + 1 - 9
y = (1/4)x^2 + x - 8

Let's double check this:
(4, 0) --> (1/4)(4)^2 + 4 - 8 --> 4 + 4 - 8 = 0
(-8, 0) --> (1/4)(-8)^2 - 8 - 8 --> 16 - 8 - 8 = 0
(0, -8) --> (1/4)(0)^2 + 0 - 8 --> 0 + 0 - 8 = -8
This all checks, so the final equation is:

Vertex form --> y + 9 = 1/4(x + 2)^2
Quadratic form --> y = (1/4)x^2 - x - 8

Either of the forms is valid, depending which you are expected to use.

Answer 2

set the equation as y=(x-a)^2+b
∴a=(x1+x2)/2 = (-8+4)/2 =-2
∴the equation becomes y=(x+2)^2+b
when x=4
(4+2) ^2+b=-8
　　　　b=-8-36
　　　　b=-44
∴the equation is
y=(x+2)^2-44

Answer 3

x interceps are -8 and 4
therefor two roots are (x+8) and (x-4)
multiply them together and you will find the equation