5.3 Graphs of polynomial functions  (Page 6/13)

As a start, evaluate $f(x)$ at the integer values $x=1,2,3,$ and $4.$ See [link] .

We see that one zero occurs at $x=2.$ Also, since $f(3)$ is negative and $f(4)$ is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.

We have shown that there are at least two real zeros between $x=1$ and $x=4.$

This graph has three x -intercepts: $x=\mathrm{-3},2,$ and $5.$ The y -intercept is located at $(0,2).$ At $x=\mathrm{-3}$ and $x=5,$ the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At $x=2,$ the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us

To determine the stretch factor, we utilize another point on the graph. We will use the $y\text{-}$ intercept $(0,–2),$ to solve for $a.$

The graphed polynomial appears to represent the function $f(x)=\frac{1}{30}(x+3){(x-2)}^2(x-5).$