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Sketch a graph of $\text{\hspace{0.17em}}f(x)=\mathrm{-2}{(x+3)}^{2}(x-5).$
This graph has two x -intercepts. At $\text{\hspace{0.17em}}x=\mathrm{-3},\text{\hspace{0.17em}}$ the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x -intercept. At $\text{\hspace{0.17em}}x=5,\text{\hspace{0.17em}}$ the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.
The y -intercept is found by evaluating $\text{\hspace{0.17em}}f(0).$
The y -intercept is $\text{\hspace{0.17em}}(0,90).$
Additionally, we can see the leading term, if this polynomial were multiplied out, would be $\text{\hspace{0.17em}}-2{x}^{3},\text{\hspace{0.17em}}$ so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See [link] .
To sketch this, we consider that:
At $\text{\hspace{0.17em}}\left(0,90\right),\text{\hspace{0.17em}}$ the graph crosses the y -axis at the y -intercept. See [link] .
Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at $\text{\hspace{0.17em}}\left(5,0\right).\text{\hspace{0.17em}}$ See [link] .
As $\text{\hspace{0.17em}}x\to \infty \text{\hspace{0.17em}}$ the function $\text{\hspace{0.17em}}f(x)\to \mathrm{-\infty},\text{\hspace{0.17em}}$ so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.
Using technology, we can create the graph for the polynomial function, shown in [link] , and verify that the resulting graph looks like our sketch in [link] .
Sketch a graph of $\text{\hspace{0.17em}}f(x)=\frac{1}{4}x{(x-1)}^{4}{(x+3)}^{3}.$
In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x -axis, we can confirm that there is a zero between them. Consider a polynomial function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ in the domain of $\text{\hspace{0.17em}}f,$ if $\text{\hspace{0.17em}}a<b\text{\hspace{0.17em}}$ and $f\left(a\right)\ne f\left(b\right),$ then the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ takes on every value between $\text{\hspace{0.17em}}f\left(a\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(b\right).\text{\hspace{0.17em}}$ (While the theorem is intuitive, the proof is actually quite complicated and requires higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ lies above the $\text{\hspace{0.17em}}x\text{-}$ axis and another point at $\text{\hspace{0.17em}}x=b\text{\hspace{0.17em}}$ lies below the $\text{\hspace{0.17em}}x\text{-}$ axis, there must exist a third point between $\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=b\text{\hspace{0.17em}}$ where the graph crosses the $\text{\hspace{0.17em}}x\text{-}$ axis. Call this point $\text{\hspace{0.17em}}\left(c,\text{}f\left(c\right)\right).\text{\hspace{0.17em}}$ This means that we are assured there is a solution $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ where $f\left(c\right)=0.$
In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the $\text{\hspace{0.17em}}x\text{-}$ axis. [link] shows that there is a zero between $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$
Let $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ be a polynomial function. The Intermediate Value Theorem states that if $\text{\hspace{0.17em}}f\left(a\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(b\right)\text{\hspace{0.17em}}$ have opposite signs, then there exists at least one value $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ between $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}f\left(c\right)=0.$
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