# 2.4: Graphs of Polynomials Using Zeros

Difficulty Level: At Grade Created by: CK-12
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Practice Graphs of Polynomials Using Zeros

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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?

### 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.

Embedded Video:

### Guidance

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 \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.

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}-5x-6\end{align*}

Solution:

Start by factoring:

\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.

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

This gives

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

Solution

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 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(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.

In the introduction to the lesson, it was noted that there are similarities in graphing using zeroes between quadratic functions and higher-degree polynomials. Were you able to identify some of those similarities?

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.

### Vocabulary

Cubic Function: A function containing an \begin{align*}x^{3}\end{align*} term as the highest power of x.

Quartic Function: A function containing an \begin{align*}x^{4}\end{align*} term as the highest power of x.

Zeroes of a Polynomial: The values output by the function (the values of \begin{align*}f(x)\end{align*} or \begin{align*}y\end{align*}), when the input (the \begin{align*}x\end{align*} value) is zero, or vice-versa.

Interval: A portion of a function, generally defined by a starting and ending value of \begin{align*}x\end{align*}.

### Guided Practice

Questions

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

Solutions

1) Solution:
Step 1: By applying the leading term test, we can say that since the co-efficient \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) Solution:
Step 1: By applying the leading term test, we can say that since the co-efficient \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) Solution:
Step 1: By applying the leading term test, we can say that since the co-efficient \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^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.

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*}.

### Practice

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 #s 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: a) The leading term: b) The degree of the polynomial: c) The leading coefficient:

Determine the equation of the polynomial based on the graph:

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### Vocabulary Language: English

TermDefinition
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.
Leading-Term Test 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.

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