# parabolas

posted by
**Emma** on
.

how do i find the equaion of the following parabolas?

1)intercepts at (2,0),(8,0) and (0,48)

2)turning point at (1,-2) and passing through the origin.

can someone please teach me how to do this??thanks!

The first: Write you standard equation of a parabola (it has two x intercepts)

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

You have three points.

Put those in the equations, you then have three equations, solve for h, b, k.

Example (first point)

0=h(2-b)^2 +k

second point

0=h(8-b)^2 + k

Third point

48=hb^2 + k

solve for h, b k There will be a lot of algebra here. A trick is to note it is symetrically about x=5, so b=5

then from the first equation k=-9h then putting that in the third,

48= h25-9h or h= 4

solved.

You can also simplify matters by making use of the location of the zeroes. You know that they are at x = 2 and x = 8, so the function must be of the form:

y(x) = A (x-2)(x-8)

To find A you use that y(0) = 48 --->

A = 3

Suppose you are given three points (x1,y1), (x2,y2) and (x3,y3) and y1, y2 and y3 are not zero. Then this trick doesn't work. However, you can still make this trick work as follows. Instead of the function y(x) you consider the function z(x) defined s:

z(x) = y(x) - y1.

Then z(x1) = 0, z(x2) = y2-y1 and

z(x3)=y3-y1

Because z(x) is zero at x = x1, you know that it contains a factor (x-x1). If you divide z(x) by (x-x1) you'll get a linear function. So, we define:

h(x) = z(x)/(x-x1).

Then h(x2)= (y2-y1 )/(x2-x1)

and

h(x3)= (y3-y1 )/(x3-x1)

h(x) is linear, you know it's value at two points, so you can easily determine it.

good post.