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Find the x -intercepts of $\text{\hspace{0.17em}}f(x)={x}^{6}-3{x}^{4}+2{x}^{2}.$
We can attempt to factor this polynomial to find solutions for
$\text{\hspace{0.17em}}f\left(x\right)=0.$
This gives us five $\text{\hspace{0.17em}}x\text{-}$ intercepts: $\text{\hspace{0.17em}}(0,0),(1,0),(-1,0),(\sqrt{2},0),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}(-\sqrt{2},0).\text{\hspace{0.17em}}$ See [link] . We can see that this is an even function.
Find the $\text{\hspace{0.17em}}x\text{-}$ intercepts of $\text{\hspace{0.17em}}f(x)={x}^{3}-5{x}^{2}-x+5.$
Find solutions for
$\text{\hspace{0.17em}}f(x)=0\text{\hspace{0.17em}}$ by factoring.
There are three $\text{\hspace{0.17em}}x\text{-}$ intercepts: $\text{\hspace{0.17em}}(-1,0),(1,0),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,0\right).\text{\hspace{0.17em}}$ See [link] .
Find the $\text{\hspace{0.17em}}y\text{-}$ and x -intercepts of $\text{\hspace{0.17em}}g(x)={(x-2)}^{2}(2x+3).$
The y -intercept can be found by evaluating $\text{\hspace{0.17em}}g\left(0\right).$
So the y -intercept is $\text{\hspace{0.17em}}(0,12).$
The x -intercepts can be found by solving $\text{\hspace{0.17em}}g\left(x\right)=0.$
So the $\text{\hspace{0.17em}}x\text{-}$ intercepts are $\text{\hspace{0.17em}}(2,0)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-\frac{3}{2},0\right).$
Find the $\text{\hspace{0.17em}}x\text{-}$ intercepts of $\text{\hspace{0.17em}}h(x)={x}^{3}+4{x}^{2}+x-6.$
This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.
Looking at the graph of this function, as shown in [link] , it appears that there are x -intercepts at $\text{\hspace{0.17em}}x=\mathrm{-3},\mathrm{-2},\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}1.$
We can check whether these are correct by substituting these values for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and verifying that
Since $\text{\hspace{0.17em}}h(x)={x}^{3}+4{x}^{2}+x-6,\text{\hspace{0.17em}}$ we have:
Each $\text{\hspace{0.17em}}x\text{-}$ intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.
Find the $\text{\hspace{0.17em}}y\text{-}$ and x -intercepts of the function $\text{\hspace{0.17em}}f(x)={x}^{4}-19{x}^{2}+30x.$
y -intercept $\text{\hspace{0.17em}}(0,0);\text{\hspace{0.17em}}$ x -intercepts $\text{\hspace{0.17em}}(0,0),(\u20135,0),(2,0),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}(3,0)$
Graphs behave differently at various $\text{\hspace{0.17em}}x\text{-}$ intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and bounce off.
Suppose, for example, we graph the function
Notice in [link] that the behavior of the function at each of the $\text{\hspace{0.17em}}x\text{-}$ intercepts is different.
The $\text{\hspace{0.17em}}x\text{-}$ intercept $\text{\hspace{0.17em}}\mathrm{-3}\text{\hspace{0.17em}}$ is the solution of equation $\text{\hspace{0.17em}}(x+3)=0.\text{\hspace{0.17em}}$ The graph passes directly through the $\text{\hspace{0.17em}}x\text{-}$ intercept at $\text{\hspace{0.17em}}x=\mathrm{-3.}\text{\hspace{0.17em}}$ The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.
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