# Need help finding a function

• May 5th 2010, 11:35 PM
mortalcyrax
Need help finding a function
I need to make a function using simultaneous equations that meet these requirements. It must go through the point of (3,0) and (6.25,10) and have x values of 9, -9 and -10 and the function cannot be a linear or quadratic.

I'm having trouble because there does not seem to be enough information given to work it out because the only points we know are (3,0) and (6.25,10).

Any help would be greatly appreciated.
• May 6th 2010, 01:58 AM
HallsofIvy
Quote:

Originally Posted by mortalcyrax
I need to make a function using simultaneous equations that meet these requirements. It must go through the point of (3,0) and (6.25,10) and have x values of 9, -9 and -10 and the function cannot be a linear or quadratic.

I'm having trouble because there does not seem to be enough information given to work it out because the only points we know are (3,0) and (6.25,10).

Any help would be greatly appreciated.

What does "have x values of 9, -9, and -10" mean? Simply that those x values are in the domain?

If that and the fact that the graph contains (3, 0) and (6.25, 10), there are an infinite number of such functions. Obviously, there exist a unique linear function through those points but since you say "the function cannot be linear or quadratic", you could just try a cubic: $y= ax^3+ bx^2+ cx+ d$. You could pick any two of a, b, c, d arbitrarily and use the two conditions to solve for the other two.

Or you could use a sin function: Take y= a sin(x)+ b. Then you must have a sin(3)+ b= 0 and a sin(6.25)+ b= 10. Solve those for a and b.
• May 6th 2010, 02:50 AM
mortalcyrax
Quote:

What does "have x values of 9, -9, and -10" mean? Simply that those x values are in the domain?
The function must past through these points.
I also forgot to mention that it must travel through y=9,-9 and -10 between x=3 and 6.25. It also must not reach y=10 until x=6.25...

I have decided to take the sin approach.

$a sin(3)+ b= 0$ Equation 1
$a sin(6.25)+ b= 10$ Equation 2

Subtract Equation 1 from 2 to eliminate b
$-0.057a=-10$

Divide both sides by -0.057 to solve for a
$a=176.9$

Sub a back into equation 2 to solve for b
$176.9 sin(6.25)+ b= 10$
$19.26+ b= 10$

Divide both sides by 19.26 to solve for b
$b=0.52$
• May 7th 2010, 09:24 PM
mortalcyrax
So is it possible to do a simultaneous equation with the information provided?