# Thread: Finding function of graph

1. ## Finding function of graph

Graph of a 3 degree polynomial is like the figure. What will be the function?

2. From the figure we see that -3 is a root, 2 is a double root, and the y-intercept is (0,10).
So the cubic polynomial has the form $f(x)=a(x+3)(x-2)^2$ for some constant $a$.
See if you can find $a$.

3. Thanks but
1-How did you find that (x+3)(x-2)^2 and not (x-2)(x+3)^2
2- You see in the figure f(0)=10 but in this solution f(0)=0

4. Hello, hejmh!

Graph of a 3 degree polynomial is like the figure.
What will be the function?

The general cubic function is: . $f(x) \:=\:ax^3 + bx^2 + cx + d$

. . Its derivative is: . $f'(x) \;=\;3ax^2 + 2bx + c$

We need to determine the four coefficients: . $a,b,c,d.$

We know five facts about the function:

. . $\begin{Bmatrix} f(0) &=& 10 \\ f(2) &=& 0 \\ f(\text{-}3) &=& 0 \\ \\[-3mm] f'(0) &=& 0 \\ f'(2) &=& 0 \end{Bmatrix}$

Set up a system of equations and solve.

Edit: Ah, I see that DrSteve has a nicer form of the cubic.

5. Originally Posted by hejmh
Thanks but
1-How did you find that (x+3)(x-2)^2 and not (x-2)(x+3)^2
Because, as Dr. Steve said, x= 2 is a double root- the x-axis is tangent to the graph at x= 2.
Another way of seeing the same thing- moving from one side of 2 to the other, say from x= 1.9 to x= 2.1, changes the sign of x- 2. x- 2 must be to an even power since y does not change sign.

2- You see in the figure f(0)=10 but in this solution f(0)=0
NO, it isn't. $f(0)= a(0+ 3)(0- 2)^2= 12a$, not 0.

6. Hello Soroban
If we set f(0)=10 then d=10
If we set f'(0)=0 then c=0
If we set f(2)=0 then 8a+4b+10=0
If we set f(-3)=0 then -27a+9b+10=0
If we set f'(2)= 0 then 12a+4b=0
Looking to tree last equations we have not unique value for a and b.

7. Originally Posted by Soroban
Hello, hejmh!

The general cubic function is: . $f(x) \:=\:ax^3 + bx^2 + cx + d$

. . Its derivative is: . $f'(x) \;=\;3ax^2 + 2bx + c$

We need to determine the four coefficients: . $a,b,c,d.$

We know five facts about the function:

. . $\begin{Bmatrix} f(0) &=& 10 \\ f(2) &=& 0 \\ f(\text{-}3) &=& 0 \\ \\[-3mm] f'(0) &=& 0 \\ f'(2) &=& 0 \end{Bmatrix}$

Set up a system of equations and solve.

Edit: Ah, I see that DrSteve has a nicer form of the cubic.
I'm not sure if it's clear from this picture that f'(0)=0. Is there a reason we can be certain of this?

8. Originally Posted by DrSteve
I'm not sure if it's clear from this picture that f'(0)=0. Is there a reason we can be certain of this?
It is like a change in direction at 0 so I think f'0) could be0.

9. According to your graph $f'(0)$ may or may not be $0$. It turns out that it is not. If you combine my post above with HallsofIvy's you will see that

$f(x)=\frac{5}{6}(x+3)(x-2)^2$.

The derivative of this function is not 0 at $x=0$.

Soroban's method should work if you just leave out the condition $f'(0)=0$. You will get a system of 3 equations in 3 unknowns which should have a unique solution (I'm pretty sure that the 3 equations in 2 unknowns you have above is inconsistent). You'll probably want to do Gauss-Jordan Reduction to solve this. It's quite a bit more work than the original solution I posted.