How can I find the cubic function of two points. I have the $y$-intersect $(0,2)$ and the $y=0$ intersect with the x-axis $(4,0)$. The equation should have the form $y=x^3+2$. But when I try to calculate the intersect with the $x$ axis by using the cubic root of $2$ I get $1.2599$. But using a plotting program the true value should be much smaller.
It's been a while since school math. What is my thinking mistake?
Edit: Corrected $y$-intersect point
Thanks to the answer the curve is:
In order to completely define a cubic function, you need four distinct points. So it is not very surprising to me that you are getting these conflicting results, since two points leaves a great deal of ambiguity.
That said, if you know in advance your equation looks like $y=ax^3+b$, then two points is enough to specify it. [Note: when I say "looks like" this I mean it has no $x$ or $x^2$ terms].
The general method for finding the coefficients in problems like this is to plug in the points for $x$ and $y$. This leaves a linear system of equations in the coefficients, which is then solvable by standard methods (like substitution or elimination).
In your case, we have the $y$-intercept $(0,2)$ and the root $(4,0)$, which gives $$0=a(4)^3+b$$ $$2=a(0)^3+b$$
The second equation immediately gives $b=2$, and then by substituting into the first we see that $0=64a+2$ and so $a=-\frac{1}{32}$.
I'm not sure where you went wrong precisely because you didn't show your reasoning all the way through [you did write the $y$-intercept as $(2,0)$ instead of $(0,2)$, though I don't think that was the real problem]. But hopefully with this solution you can diagnose the problem yourself :)
Either your equation is wrong or your $x$-axis intercept is wrong. If the equation is right, the intercept should be $( - \sqrt[3]{2},0)$.
