Question

1. a function f(x) has exactly three real zeros at -3, 3, and 4. the graph of the function f(x) has a y - intercept of 36 and the following end behavior: x → ∞, f(x) → ∞ and x → - ∞, f(x) → - ∞. write a possible function, you don’t need to foil it.

Explanation:

Step1: Use zero - factor form

If \(x = a\) is a zero of a function, then \((x - a)\) is a factor. Since the zeros are \(x=-3,3,4\), the function has factors \((x + 3)\), \((x - 3)\) and \((x - 4)\). So a general form of the function is \(f(x)=a(x + 3)(x - 3)(x - 4)\).

Step2: Find the value of \(a\)

The \(y\) - intercept is found by setting \(x = 0\). We know that \(f(0)=36\). Substitute \(x = 0\) into \(f(x)=a(x + 3)(x - 3)(x - 4)\):
\[

$$\begin{align*} f(0)&=a(0 + 3)(0 - 3)(0 - 4)\\ 36&=a(3)(-3)(-4)\\ 36&=a(36)\\ a& = 1 \end{align*}$$

\]

Answer:

\(f(x)=(x + 3)(x - 3)(x - 4)\)