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# Graphs of Polynomials Using Zeros

## x- and y-intercepts, signs, and leading term tests used to sketch a graph.

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Graphs of Polynomials Using Zeros

How is finding and using the zeroes of a higher-degree polynomial related to the same process you have used in the past on quadratic functions?

### Graphing Polynomials Using Zeros

The following procedure can be followed when graphing a polynomial function.

• Use the leading-term test to determine the end behavior of the graph.
• Find the x\begin{align*}x-\end{align*}intercept(s) of f(x)\begin{align*}f(x)\end{align*} by setting f(x)=0\begin{align*}f(x)=0\end{align*} and then solving for x\begin{align*}x\end{align*}.
• Find the y\begin{align*}y-\end{align*}intercept of f(x)\begin{align*}f(x)\end{align*} by setting y=f(0)\begin{align*}y=f(0)\end{align*} and finding y\begin{align*}y\end{align*}.
• Use the x\begin{align*}x-\end{align*}intercept(s) to divide the x\begin{align*}x-\end{align*}axis into intervals and then choose test points to determine the sign of f(x)\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.

If anxn\begin{align*}a_{n}x^{n}\end{align*} is the leading term of a polynomial. Then the behavior of the graph as x\begin{align*}x\to\infty\end{align*} or x\begin{align*}x\to-\infty\end{align*} can be known by one the four following behaviors:

1. If an>0\begin{align*}a_{n}>0\end{align*} and n\begin{align*}n\end{align*} even:

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

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

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

### Examples

#### Example 1

Earlier, you were asked to identify some similarities in graphing using zeroes between quadratic functions and higher-degree polynomials.

Despite the more complex nature of the graphs of higher-degree 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.

#### Example 2

Find the roots (zeroes) of the polynomial:

h(x)=x3+2x25x6\begin{align*}h(x)=x^{3}+2x^{2}-5x-6\end{align*}

Start by factoring:

h(x)=x3+2x25x6=(x+1)(x2)(x+3)\begin{align*}h(x)=x^{3}+2x^{2}-5x-6=(x+1)(x-2)(x+3)\end{align*}

To find the zeros, set h(x)=0 and solve for x.

(x+1)(x2)(x+3)=0\begin{align*}(x+1)(x-2)(x+3)=0\end{align*}

This gives

x+1x2x+3=0=0=0\begin{align*}x+1 & = 0\\ x-2 & = 0\\ x+3 & = 0\end{align*}

or

xxx=1=2=3\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 3

Find the zeros of \begin{align*}g(x)=-(x-2)(x-2)(x+1)(x+5)(x+5)(x+5)\end{align*}.

The polynomial can be written as

\begin{align*}g(x)=-(x-2)^{2}(x+1)(x+5)^{3}\end{align*}

To solve the equation, we simply set it equal to zero

\begin{align*}-(x-2)^{2}(x+1)(x+5)^{3}=0\end{align*}

this gives

\begin{align*}x-2 & = 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*}(x-2)\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*}(x-2)\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 4

Graph the polynomial function \begin{align*}f(x)=-3x^{4}+2x^{3}\end{align*}.

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(3x-2) & = 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.

#### Example 5

Find the zeros and sketch a graph of the polynomial

\begin{align*}f(x)=x^{4}-x^{2}-56\end{align*}

This is a factorable equation,

\begin{align*}f(x) & =x^4-x^2-56\\ & = (x^2-8)(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.

#### Example 6

Graph \begin{align*}g(x)=-(x-2)^{2}(x+1)(x+5)^{3}\end{align*}.

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*}.

### Review

1. If c is a zero of f, then c is a/an _________________________ of the graph of f.
2. If c is a zero of f, then (x - c) is a factor of ___________________?
3. 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*}.

1. How many zeros (x-intercepts) are there?
2. What is the leading term?

Find the zeros and graph the polynomial. Be sure to label the x-intercepts, y-intercept (if possible) and have correct end behavior. You may use technology for questions 9-12.

1. \begin{align*}P(x) = -2(x + 1)^2(x - 3)\end{align*}
2. \begin{align*}P(x) = x^3 + 3x^2 - 4x - 12\end{align*}
3. \begin{align*}f(x) = -2x^3 + 6x^2 + 9x + 6\end{align*}
4. \begin{align*}f(x) = -4x^2 -7x +3\end{align*}
5. \begin{align*}f(x) = 2x^5 +4x^3 + 8x^2 +6x\end{align*}
6. \begin{align*}f(x) = x^4 - 3x^2\end{align*}
7. \begin{align*}g(x) = x^2 - |x|\end{align*}
8. Given: \begin{align*}P(x) = (3x +2)(x - 7)^2(9x + 2)^3\end{align*}

State:

2. The degree of the polynomial:

Determine the equation of the polynomial based on the graph:

To see the Review answers, open this PDF file and look for section 2.4.

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Color Highlighted Text Notes

### Vocabulary Language: English

Cubic Function

A cubic function is a function containing an $x^{3}$ term as the highest power of $x$.

Intercept

The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.

interval

An interval is a specific and limited part of a function.

The leading-term test is a test to determine the end behavior of a polynomial function.

Polynomial

A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.

Polynomial Graph

A polynomial graph is the graph of a polynomial function. The term is most commonly used for polynomial functions with a degree of at least three.

Quartic Function

A quartic function is a function $f(x)$ containing an $x^{4}$ term as the highest power of ''x''.

Roots

The roots of a function are the values of x that make y equal to zero.

Zeroes

The zeroes of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.