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Access the following online resource for additional instruction and practice with graphing polynomial functions.
What is the difference between an $\text{\hspace{0.17em}}x\text{-}$ intercept and a zero of a polynomial function $\text{\hspace{0.17em}}f?\text{\hspace{0.17em}}$
The $\text{\hspace{0.17em}}x\text{-}$ intercept is where the graph of the function crosses the $\text{\hspace{0.17em}}x\text{-}$ axis, and the zero of the function is the input value for which $\text{\hspace{0.17em}}f(x)=0.$
If a polynomial function of degree $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ has $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ distinct zeros, what do you know about the graph of the function?
Explain how the Intermediate Value Theorem can assist us in finding a zero of a function.
If we evaluate the function at $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and the sign of the function value changes, then we know a zero exists between $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.$
Explain how the factored form of the polynomial helps us in graphing it.
If the graph of a polynomial just touches the x -axis and then changes direction, what can we conclude about the factored form of the polynomial?
There will be a factor raised to an even power.
For the following exercises, find the $\text{\hspace{0.17em}}x\text{-}$ or t -intercepts of the polynomial functions.
$\text{\hspace{0.17em}}C\left(t\right)=2\left(t-4\right)\left(t+1\right)(t-6)$
$\text{\hspace{0.17em}}C\left(t\right)=3\left(t+2\right)\left(t-3\right)(t+5)$
$(-2,0),(3,0),(-5,0)$
$\text{\hspace{0.17em}}C\left(t\right)=4t{\left(t-2\right)}^{2}(t+1)$
$\text{\hspace{0.17em}}C\left(t\right)=2t\left(t-3\right){\left(t+1\right)}^{2}$
$\text{\hspace{0.17em}}(3,0),(-1,0),(0,0)$
$\text{\hspace{0.17em}}C\left(t\right)=2{t}^{4}-8{t}^{3}+6{t}^{2}$
$\text{\hspace{0.17em}}C\left(t\right)=4{t}^{4}+12{t}^{3}-40{t}^{2}$
$\left(0,0\right),\text{}\left(-5,0\right),\text{}\left(2,0\right)$
$\text{\hspace{0.17em}}f(x)={x}^{4}-{x}^{2}$
$\text{\hspace{0.17em}}f(x)={x}^{3}+{x}^{2}-20x$
$\left(0,0\right),\text{}\left(-5,0\right),\text{}\left(4,0\right)$
$f(x)={x}^{3}+6{x}^{2}-7x$
$f(x)={x}^{3}+{x}^{2}-4x-4$
$\left(2,0\right),\text{}\left(-2,0\right),\text{}\left(-1,0\right)$
$f(x)={x}^{3}+2{x}^{2}-9x-18$
$f(x)=2{x}^{3}-{x}^{2}-8x+4$
$(-2,0),\text{\hspace{0.17em}}(2,0),\text{\hspace{0.17em}}\left(\frac{1}{2},0\right)$
$f(x)={x}^{6}-7{x}^{3}-8$
$f(x)=2{x}^{4}+6{x}^{2}-8$
$\left(1,0\right),\text{}\left(-1,0\right)$
$f(x)={x}^{3}-3{x}^{2}-x+3$
$f(x)={x}^{6}-2{x}^{4}-3{x}^{2}$
$(0,0),\text{\hspace{0.17em}}(\sqrt{3},0),\text{\hspace{0.17em}}(-\sqrt{3},0)$
$f(x)={x}^{6}-3{x}^{4}-4{x}^{2}$
$f(x)={x}^{5}-5{x}^{3}+4x$
$\left(0,0\right),\text{}\left(1,0\right)\text{,}\left(-1,0\right),\text{}\left(2,0\right),\text{}\left(-2,0\right)$
For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.
$f(x)={x}^{3}-9x,\text{\hspace{0.17em}}$ between $\text{\hspace{0.17em}}x=\mathrm{-4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=\mathrm{-2.}$
$f(x)={x}^{3}-9x,\text{\hspace{0.17em}}$ between $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=4.$
$f\left(2\right)=\u201310\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(4\right)=28.$ Sign change confirms.
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