# Find Equation Of Parabola

• Apr 8th 2007, 05:44 AM
r_maths
Find Equation Of Parabola
I just need to know, how to find equation of a parabola and put it in the form of ax^2+bx+c just from reading of the graph.

I can do this for graphs like these were it cuts the x-axis:
http://img141.imageshack.us/img141/7839/surf006lm9.jpg

but for graphs were it doesn't cut the x-axis:
http://img141.imageshack.us/img141/2050/surf007hz9.jpg
how would you find the equation?

Thanks
• Apr 8th 2007, 06:47 AM
Soroban
Hello, r_maths!

Quote:

I just need to know, how to find equation of a parabola
and put it in the form of ax² + bx + c just from reading the graph.

I can do this for graphs where it cuts the x-axis.
But for graphs were it doesn't cut the x-axis,
how would you find the equation?

Code:

```          |         *|                    *           |         A o(0,18)          C o(4,18)           | *              *           |    *    B    *           |        o           |      (2,10)           |           |       - - + - - - - - - - - - - -           |```

I assume that point B(2,10) is the vertex of the parabola.

The parabola is symmetric to its axis: x = 2.
. . Hence, there is a point C(4,18) symmetric to A(0,18).

The parabola has the general equation: .y .= .ax² + bx + c

We can evaulate these coefficients, using the three given points.

(0,18): .a·0² + b·0 + c .= .18 . . c = 18

(2,10): .a·2² + b·2 + 18 .= .10 . . 4a + 2b .= .-8 .[1]

(4,18): .a·4² + b·4 + 18 .= .18 . . 16a + 4b .= .0 .[2]

Multiply [1] by -2: . -8a - 4b .= .16
. . - . . . .Add [2]: .16a + 4b .= .0

And we get: .8a = 16 . . a = 2

Substitute into [1]: .8 + 2b .= -8 . .b = -8

Therefore: .y .= .2x² - 8x + 18

• Apr 8th 2007, 07:46 AM
r_maths
Thanks Soroban