Answer:
E. f(x)= (x+10)(x-3)^2 (x+1)
Explanation:
We can use the equations to find the roots. As the graph shows that there are roots at x=-10, x=-1, and x=3, we can use this to find the equation.
We know that if there is a root, one of the values have to equal 0. Therefore, the equation is (x+10)(x-3)(x+1) (as if x+10=0, then x=-10, which fulfills our requirement from above. The same goest with (x-3) and x=3 and (x+1) and x=-1.)
However looking at the graph we see that f(x) does not pass through x=3 but rather just touches it. When a function touches a root, we know that the exponent of the value is even. When the function passes through the root, we know that the exponent of the value is odd.
Taking this into account, the only one that fulfills all these requirements is E, as it has (x+10)(x-3)(x+1) and the exponent of (x-3) is even so that the function bounces rather than passes through.