How is finding and using the zeroes of a higherdegree polynomial related to the same process you have used in the past on quadratic functions?
Watch This
This video is a good introduction to graphing cubic and higher degree polynomials. Note that a graphing calculator is used in the video.
James Sousa: Graphing Cubic Functions
Guidance
The following procedure can be followed when graphing a polynomial function.
 Use the leadingterm test to determine the end behavior of the graph.
 Find the \begin{align*}x\end{align*}intercept(s) of \begin{align*}f(x)\end{align*} by setting \begin{align*}f(x)=0\end{align*} and then solving for \begin{align*}x\end{align*}.
 Find the \begin{align*}y\end{align*}intercept of \begin{align*}f(x)\end{align*} by setting \begin{align*}y=f(0)\end{align*} and finding \begin{align*}y\end{align*}.
 Use the \begin{align*}x\end{align*}intercept(s) to divide the \begin{align*}x\end{align*}axis into intervals and then choose test points to determine the sign of \begin{align*}f(x)\end{align*} on each interval.
 Plot the test points.
 If necessary, find additional points to determine the general shape of the graph.
The LeadingTerm Test
If \begin{align*}a_{n}x^{n}\end{align*} is the leading term of a polynomial. Then the behavior of the graph as \begin{align*}x\to\infty\end{align*} or \begin{align*}x\to\infty\end{align*} can be known by one the four following behaviors:
1. If \begin{align*}a_{n}>0\end{align*} and \begin{align*}n\end{align*} even: 

2. If \begin{align*}a_{n}<0\end{align*} and \begin{align*}n\end{align*} even: 


3. If \begin{align*}a_{n}>0\end{align*} and \begin{align*}n\end{align*} odd: 

4. If \begin{align*}a_{n}<0\end{align*} and \begin{align*}n\end{align*} odd: 


Example A
Find the roots (zeroes) of the polynomial:
\begin{align*}h(x)=x^{3}+2x^{2}5x6\end{align*}
Solution:
Start by factoring:
\begin{align*}h(x)=x^{3}+2x^{2}5x6=(x+1)(x2)(x+3)\end{align*}
To find the zeros, set h(x)=0 and solve for x.
\begin{align*}(x+1)(x2)(x+3)=0\end{align*}
This gives
\begin{align*}x+1 & = 0\\ x2 & = 0\\ x+3 & = 0\end{align*}
or
\begin{align*}x & = 1\\ x & = 2\\ x & = 3\end{align*}
So we say that the solution set is \begin{align*}\{3, 1, 2\}\end{align*}. They are the zeros of the function \begin{align*}h(x)\end{align*}. The zeros of \begin{align*}h(x)\end{align*} are the \begin{align*}x\end{align*}intercepts of the graph \begin{align*}y=h(x)\end{align*} below.
Example B
Find the zeros of \begin{align*}g(x)=(x2)(x2)(x+1)(x+5)(x+5)(x+5)\end{align*}.
Solution:
The polynomial can be written as
\begin{align*}g(x)=(x2)^{2}(x+1)(x+5)^{3}\end{align*}
To solve the equation, we simply set it equal to zero
\begin{align*}(x2)^{2}(x+1)(x+5)^{3}=0\end{align*}
this gives
\begin{align*}x2 & = 0\\ x+1 & = 0\\ x+5 & = 0\end{align*}
or
\begin{align*}x & = 2\\ x & = 1\\ x & = 5\end{align*}
Notice the occurrence of the zeros in the function. The factor \begin{align*}(x2)\end{align*} occurred twice (because it was squared), the factor \begin{align*}(x+1)\end{align*} occurred once and the factor \begin{align*}(x+5)\end{align*} occurred three times. We say that the zero we obtain from the factor \begin{align*}(x2)\end{align*} has a multiplicity \begin{align*}k=2\end{align*} and the factor \begin{align*}(x+5)\end{align*} has a multiplicity \begin{align*}k=3\end{align*}.
Example C
Graph the polynomial function \begin{align*}f(x)=3x^{4}+2x^{3}\end{align*}.
Solution:
Since the leading term here is \begin{align*}3x^{4}\end{align*} then \begin{align*}a_{n}=3<0\end{align*}, and \begin{align*}n=4\end{align*} even. Thus the end behavior of the graph as \begin{align*}x\to\infty\end{align*} and \begin{align*}x\to\infty\end{align*} is that of Box #2, item 2.
We can find the zeros of the function by simply setting \begin{align*}f(x)=0\end{align*} and then solving for \begin{align*}x\end{align*}.
\begin{align*}3x^{4}+2x^{3} & = 0\\ x^3(3x2) & = 0\end{align*}
This gives
\begin{align*}x=0\quad \text{or} \quad x=\frac{2}{3}\end{align*}
So we have two \begin{align*}x\end{align*}intercepts, at \begin{align*}x=0\end{align*} and at \begin{align*}x=\frac{2}{3}\end{align*}, with multiplicity \begin{align*}k=3\end{align*} for \begin{align*}x=0\end{align*} and multiplicity \begin{align*}k=1\end{align*} for \begin{align*}x=\frac{2}{3}\end{align*}.
To find the \begin{align*}y\end{align*}intercept, we find \begin{align*}f(0)\end{align*}, which gives
\begin{align*}f(0)=0\end{align*}
So the graph passes the \begin{align*}y\end{align*}axis at \begin{align*}y=0\end{align*}.
Since the \begin{align*}x\end{align*}intercepts are 0 and \begin{align*}\frac{2}{3}\end{align*}, they divide the \begin{align*}x\end{align*}axis into three intervals: \begin{align*}(\infty, 0), \left ( 0, \frac{2}{3} \right ),\end{align*} and \begin{align*}\left ( \frac{2}{3}, \infty \right )\end{align*}. Now we are interested in determining at which intervals the function \begin{align*}f(x)\end{align*} is negative and at which intervals it is positive. To do so, we construct a table and choose a test value for \begin{align*}x\end{align*} from each interval and find the corresponding \begin{align*}f(x)\end{align*} at that value.
Interval  Test Value \begin{align*}x\end{align*}  \begin{align*}f(x)\end{align*}  Sign of \begin{align*}f(x)\end{align*}  Location of points on the graph 

\begin{align*}(\infty, 0)\end{align*}  1  5    below the \begin{align*}x\end{align*}axis 
\begin{align*}\left ( 0, \frac{2}{3} \right )\end{align*}  \begin{align*}\frac{1}{2}\end{align*}  \begin{align*}\frac{1}{16}\end{align*}  +  above the \begin{align*}x\end{align*}axis 
\begin{align*}\left ( \frac{2}{3}, \infty \right )\end{align*}  1  1    below the \begin{align*}x\end{align*}axis 
Those test points give us three additional points to plot: \begin{align*}(1, 5), \left ( \frac{1}{2},\frac{1}{16} \right )\end{align*}, and (1, 1). Now we are ready to plot our graph. We have a total of three intercept points, in addition to the three test points. We also know how the graph is behaving as \begin{align*}x\to\infty\end{align*} and \begin{align*}x\to+\infty\end{align*}. This information is usually enough to make a rough sketch of the graph. If we need additional points, we can simply select more points to complete the graph.
Concept question wrapup: In the introduction to the lesson, it was noted that there are similarities in graphing using zeroes between quadratic functions and higherdegree polynomials. Were you able to identify some of those similarities? Despite the more complex nature of the graphs of higherdegree polynomials, the general process of graphing using zeroes is actually very similar. In both cases, your goal is to locate the points where the graph crosses the x or y axis. In both cases, this is done by setting the y value equal to zero and solving for x to find the x axis intercepts, and setting the x value equal to zero and solving for y to find the y axis intercepts. 

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Guided Practice
Sketch a graph of each power function using the properties of the power functions.
1) \begin{align*}f(x)=3x^{4}\end{align*}
2) \begin{align*}h(x)=\frac{1}{2}x^{5}\end{align*}
3) \begin{align*}q(x)=4x^{8}\end{align*}
4) Find the zeros and sketch a graph of the polynomial
\begin{align*}f(x)=x^{4}x^{2}56\end{align*}
5) Graph \begin{align*}g(x)=(x2)^{2}(x+1)(x+5)^{3}\end{align*}
Answers
 1) Step 1: By applying the leading term test, we can say that since the coefficient \begin{align*}3\end{align*} is \begin{align*} <0\end{align*}, and since the power \begin{align*}4\end{align*} is even, the end behavior of the graph resembles:

 Step 2: By solving the equation for \begin{align*}x = 1\end{align*} and \begin{align*}x = 1\end{align*}, we get the points: \begin{align*}(1, 3)\end{align*} and \begin{align*}(1, 3)\end{align*}.

 Step 3: This suggests
 2) Step 1: By applying the leading term test, we can say that since the coefficient \begin{align*}\frac{1}{2}\end{align*} is \begin{align*} >0\end{align*}, and since the power \begin{align*}5\end{align*} is odd, the end behavior of the graph resembles:

 Step 2: By solving the equation for \begin{align*}x = 1\end{align*} and \begin{align*}x = 1\end{align*}, we get the points: \begin{align*}(1, 1/2)\end{align*} and \begin{align*}(1, 1/2)\end{align*}.

 Step 3: This suggests
 3) Step 1: By applying the leading term test, we can say that since the coefficient \begin{align*}4\end{align*} is \begin{align*} >0\end{align*}, and since the power \begin{align*}8\end{align*} is even, the end behavior of the graph resembles:

 Step 2: By solving the equation for \begin{align*}x = 1\end{align*} and \begin{align*}x = 1\end{align*}, we get the points: \begin{align*}(1, 4)\end{align*} and \begin{align*}(1, 4)\end{align*}.

 Step 3: This suggests
4) This is a factorable equation,
\begin{align*}f(x) & =x^4x^256\\ & = (x^28)(x^2+7)\end{align*}
Setting \begin{align*}f(x)=0\end{align*},
\begin{align*}(x^{2}8)(x^{2}+7) = 0\end{align*}
the first term gives
\begin{align*}x^{2}8 & = 0\\ x^2 & = 8\\ x & = \pm \sqrt{8}\\ & = \pm 2\sqrt{2}\end{align*}
and the second term gives
\begin{align*}x^{2}+7 & = 0\\ x^2 & = 7\\ x & = \pm \sqrt{7}\\ & = \pm i\sqrt{7}\end{align*}
So the solutions are \begin{align*}\pm2\sqrt{2}\end{align*} and \begin{align*}\pm i\sqrt{7}\end{align*}, a total of four zeros of \begin{align*}f(x)\end{align*}. Keep in mind that only the real zeros of a function correspond to the \begin{align*}x\end{align*}intercept of its graph.
5) Use the zeros to create a table of intervals and see whether the function is above or below the \begin{align*}x\end{align*}axis in each interval:
Interval  Test value \begin{align*}x\end{align*}  \begin{align*}g(x)\end{align*}  Sign of \begin{align*}g(x)\end{align*}  Location of graph relative to \begin{align*}x\end{align*}axis 

\begin{align*}(\infty, 5)\end{align*}  6  320  +  Above 
\begin{align*}x=5\end{align*}  5  0  NA  
(5, 1)  2  144  +  Above 
\begin{align*}x=1\end{align*}  1  0  NA  
(1, 2)  0  100    Below 
\begin{align*}x=2\end{align*}  2  0  NA  
\begin{align*}(2, \infty)\end{align*}  3  256    Below 
Finally, use this information and the test points to sketch a graph of \begin{align*}g(x)\end{align*}.
Explore More
 If c is a zero of f, then c is a/an _________________________ of the graph of f.
 If c is a zero of f, then (x  c) is a factor of ___________________?
 Find the zeros of the polynomial: \begin{align*}P(x) = x^3  5x^2 + 6x\end{align*}
Consider the function: \begin{align*}f(x) = 3(x  3)^4(5x  2)(2x  1)^3(4  x)^2\end{align*}.
 How many zeros (xintercepts) are there?
 What is the leading term?
Find the zeros and graph the polynomial. Be sure to label the xintercepts, yintercept (if possible) and have correct end behavior. You may use technology for #s 912
 \begin{align*}P(x) = 2(x + 1)^2(x  3)\end{align*}
 \begin{align*}P(x) = x^3 + 3x^2  4x  12\end{align*}
 \begin{align*}f(x) = 2x^3 + 6x^2 + 9x + 6\end{align*}
 \begin{align*}f(x) = 4x^2 7x +3\end{align*}
 \begin{align*}f(x) = 2x^5 +4x^3 + 8x^2 +6x\end{align*}
 \begin{align*}f(x) = x^4  3x^2\end{align*}
 \begin{align*}g(x) = x^2  x\end{align*}
 Given: \begin{align*}P(x) = (3x +2)(x  7)^2(9x + 2)^3\end{align*} State: a) The leading term: b) The degree of the polynomial: c) The leading coefficient:
Determine the equation of the polynomial based on the graph:
Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 2.4.