# 2.2.2: Graphs of Polynomials Using Zeros

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- 14193

# 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−
**intercept**(s) of f(x) by setting f(x)=0 and then solving for x. - Find the y−intercept of f(x) by setting y=f(0) and finding y.
- Use the x−intercept(s) to divide the x−axis into intervals and then choose test points to determine the sign of f(x) on each
**interval**. - Plot the test points.
- If necessary, find additional points to determine the general shape of the graph.

# The Leading-Term Test

If a_{n}x^{n} is the leading term of a polynomial. Then the behavior of the graph as x→∞ or x→−∞ can be known by one the four following behaviors:

1. If a_{n}>0 and n even:

2. If a_{n}<0 and n even:

3. If a_{n}>0 and n odd:

4. If a_{n}<0 and n odd:

# Examples

Example 1

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

**Solution**

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

*axis. In both cases, this is done by setting the*

*y**value equal to zero and solving for*

*y**to find the*

*x**axis intercepts, and setting the*

*x**value equal to zero and solving for*

*x**to find the*

*y**axis intercepts.*

*y*Example 2

Find the **roots** (zeroes) of the polynomial:

h(x)=x^{3}+2x^{2}−5x−6

**Solution**

Start by factoring:

h(x)=x^{3}+2x^{2}−5x−6=(x+1)(x−2)(x+3)

To find the zeros, set * h*(

*)=0 and solve for*

*x**.*

*x*(x+1)(x−2)(x+3)=0

This gives

x+1=0

x−2=0

x+3=0

or

x=-1

x=2

x=-3

So we say that the solution set is {−3,−1,2}. They are the zeros of the function h(x). The zeros of h(x) are the x−intercepts of the graph y=h(x) below.

Example 3

Find the zeros of g(x)=−(x−2)(x−2)(x+1)(x+5)(x+5)(x+5).

**Solution**

The polynomial can be written as

g(x)=−(x−2)^{2}(x+1)(x+5)^{3}

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

−(x−2)^{2}(x+1)(x+5)^{3}=0

this gives

x−2=0

x+1=0

x+5=0

or

x=2

x=-1

x=-5

Notice the occurrence of the zeros in the function. The factor (x−2) occurred twice (because it was squared), the factor (x+1) occurred once and the factor (x+5) occurred three times. We say that the zero we obtain from the factor (x−2) has a * multiplicity* k=2 and the factor (x+5) has a

*k=3.*

*multiplicity*Example 4

Graph the polynomial function f(x)=−3x^{4}+2x^{3}.

**Solution**

Since the leading term here is −3x^{4} then a_{n}=−3<0, and n=4 even. Thus the end behavior of the graph as x→∞ and x→−∞ is that of Box #2, item 2.

We can find the zeros of the function by simply setting f(x)=0 and then solving for x.

−3x^{4}+2x^{3}=0

−x^{3}(3x−2)=0

This gives

x=0 or x=\(\ 2\over 3\)

So we have two x−intercepts, at x=0 and at x=\(\ 2\over 3\), with multiplicity k=3 for x=0 and multiplicity k=1 for x=\(\ 2\over 3\)

To find the y−intercept, we find f(0), which gives

f(0)=0

So the graph passes the y−axis at y=0.

Since the x−intercepts are 0 and \(\ 2\over 3\), they divide the x−axis into three intervals: (−∞, 0), (0, \(\ 2\over 3\)), and (\(\ 2\over 3\), ∞). Now we are interested in determining at which intervals the function f(x) is negative and at which intervals it is positive. To do so, we construct a table and choose a test value for x from each interval and find the corresponding f(x) at that value.

Interval |
Test Value x |
f(x) | Sign of f(x) |
Location of points on the graph |
---|---|---|---|---|

(−∞, 0) | -1 | -5 | - | below the x−axis |

(0, \(\ 2\over 3\)) | \(\ 1\over 2\) | \(\ 1\over 16\) | + | above the x−axis |

(\(\ 2\over 3\), ∞) | 1 | -1 | - | below the x−axis |

Those test points give us three additional points to plot: (−1, −5), (\(\ 1\over 2\), \(\ 1\over 16\)), 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 x→−∞ and x→+∞. 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

f(x)=x^{4}−x^{2}−56

**Solution**

This is a factorable equation,

f(x)=x^{4}−x^{2}−56

=(x^{2}−8)(x^{2}+7)

Setting f(x)=0,

(x^{2}−8)(x^{2}+7)=0

the first term gives

x^{2}−8=0

x^{2}=0

x= ±\(\ \sqrt{8}\)

= ±\(\ 2\sqrt{2}\)

and the second term gives

x^{2}+7=0

x^{2}=−7

x=±\(\ \sqrt{-7}\)

=±\(\ i\sqrt{7}\)

So the solutions are ±\(\ 2\sqrt{2}\) and ±\(\ i\sqrt{7}\), a total of four zeros of f(x). Keep in mind that only the * real* zeros of a function correspond to the x−intercept of its graph.

Example 6

Graph g(x)=−(x−2)^{2}(x+1)(x+5)^{3}.

**Solution**

Use the zeros to create a table of intervals and see whether the function is above or below the x−axis in each interval:

Interval |
Test value x |
g(x) |
Sign of g(x) |
Location of graph relative to x−axis |
---|---|---|---|---|

(−∞, −5) | -6 | 320 | + | Above |

x=−5 | -5 | 0 | NA | |

(-5, -1) | -2 | 144 | + | Above |

x=−1 | -1 | 0 | NA | |

(-1, 2) | 0 | -100 | - | Below |

x=2 | 2 | 0 | NA | |

(2, ∞) | 3 | -256 | - | Below |

Finally, use this information and the test points to sketch a graph of g(x).

# Review

- If
is a zero of*c*, then*f*is a/an _________________________ of the graph of*c*.*f* - If
is a zero of*c*, then (*f*-*x*) is a factor of ___________________?*c* - Find the zeros of the polynomial: P(x)=x
^{3}−5x^{2}+6x

Consider the function: f(x)=−3(x−3)^{4}(5x−2)(2x−1)^{3}(4−x)^{2}.

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

Find the zeros and graph the polynomial. Be sure to label the * x*-intercepts,

*-intercept (if possible) and have correct end behavior. You may use technology for questions 9-12.*

*y*- P(x)=−2(x+1)
^{2}(x−3) - P(x)=x
^{3}+3x^{2}−4x−12 - f(x)=−2x
^{3}+6x^{2}+9x+6 - f(x)=−4x
^{2}−7x+3 - f(x)=2x
^{5}+4x^{3}+8x^{2}+6x - f(x)=x
^{4}−3x^{2} - g(x)=x
^{2}−|x| - Given: P(x)=(3x+2)(x−7)
^{2}(9x+2)^{3}

State:

- The leading term:
- The degree of the polynomial:
- The leading coefficient:

Determine the equation of the polynomial based on the graph:

# Vocabulary

Term | Definition |
---|---|

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 that make x equal to zero.y |

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