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Question

I need to determine the equation of the function in the graph below.

Attached is a graph with x-intercepts at (-4, 0) and (3,0) and another point given at (2,10). I know its not quadratic. Are complex zeros involved, and if so how do I find them?

Answer 1

Your function, presumably a polynomial, seems to have roots at $-4$ and $3$, thus is divisible by $(x+4)(x-3)=x^2+x-12$. Assuming that it’s of the minimal degree to have that shape, namely degree four, you can write it as $(ax^2+bx+c)(x^2+x-12)=ax^4+(a+b)x^3+(-12a+b+c)x^2+(-12b+c)x-12c$. But it also passes through the point $(2,10)$, so you can get a relation among the undetermined coefficients $a$, $b$, and $c$, namely $-24a - 12b - 6c=10$.
More than that, it seems that your function has a local minimum at $x=0$ and a local maximum at $x=2$. So when you evaluate $f'$ at $0$ and at $2$, you’ll get $0$, and these two facts give you two more linear relations among $a$, $b$, and $c$. With the three relations, you can solve for the unknowns. I’ll leave that to you.