### About Graphs of Polynomial Functions:

In this section, we have studied the graphs of polynomial functions. We have learned about graphing transformations of monomials. Additionally, we have studied all of the characteristics that allow us to create a rough sketch of a polynomial function. This would include finding the end behavior, which tells us how the function behaves as x increases or decreases without bound. We also studied the shape of a graph near a zero. This discussion about the shape tells us whether the graph will cross the x-axis or touch the x-axis and turn around. Additionally, we have learned how to find enough ordered pairs using test points to create a somewhat accurate graph.

Test Objectives

- Demonstrate the ability to graph transformations of monomials
- Demonstrate the ability to find the end behavior of a polynomial function
- Demonstrate the ability to find the shape of the graph near a zero
- Demonstrate the ability to identify the local maxima and minima from a graph
- Demonstrate the ability to create a rough sketch of a polynomial function

#1:

Instructions: Describe the transformation from f(x) to g(x) and then sketch the graph of g(x).

$$a)\hspace{.2em}$$ $$f(x) = x^3$$ $$g(x) = \frac{1}{2}(x - 2)^3 - 1$$

$$b)\hspace{.2em}$$ $$f(x) = x^4$$ $$g(x) = \frac{1}{4}(x + 3)^4 + 1$$

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#2:

Instructions: Describe the end behavior of each polynomial function.

$$a)\hspace{.2em}$$ $$f(x) = -2x^6 + 4x^3 - x^2 + 1$$

$$b)\hspace{.2em}$$ $$f(x) = 5x^3 - x + 3$$

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#3:

Instructions: Identify the x-intercepts and describe the shape of the graph near that x-intercept.

$$a)\hspace{.2em}$$ $$f(x) = \frac{1}{10}(x - 1)^3(2x - 5)^2(x + 1)$$

$$b)\hspace{.2em}$$ $$f(x) = \frac{1}{3}(x - 2)^3(x - 5)^2(2x - 9)$$

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#4:

Instructions: Use the given function to state the maximum number of turning points and use the given graph to identify the local maximum and minimum points.

$$a)\hspace{.2em}f(x) = \frac{1}{4}x^4 - 2x^3 + \frac{11}{2}x^2 - 6x + \frac{17}{4}$$ Desmos Link for More Detail

$$b)\hspace{.2em}f(x) = -x^3 + 6x^2 - 9x + 4$$ Desmos Link for More Detail

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#5:

Instructions: Create a rough sketch of the given polynomial function.

Note: It is unnecessary to try and create a perfect graph. Just try to get it somewhat close to the computer-drawn version.

$$a)\hspace{.2em}f(x) = 2(x - 1)(x - 5)(x - 3)$$

$$b)\hspace{.2em}f(x) = (2x - 3)^2(x - 4)^2$$

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Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}$$ Vertically compressed by a factor of 2 (could also say vertically shrunk by a factor of 1/2), shifted 2 units right, and shifted 1 unit down.

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$$b)\hspace{.2em}$$ Vertically compressed by a factor of 4 (could also say vertically shrunk by a factor of 1/4), shifted 3 units left, and shifted 1 unit up.

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#2:

Solutions:

$$a)\hspace{.2em}$$ The graph falls to the left and falls to the right.

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$$b)\hspace{.2em}$$ The graph falls to the left and rises to the right.

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#3:

Solutions:

$$a)\hspace{.2em}$$ x-intercepts: $$(1, 0), \left(\frac{5}{2}, 0\right), (-1, 0)$$ The zero of 1 has multiplicity 3, which is odd. Therefore, at (1, 0), the graph will cross through the x-axis. The zero of 5/2 has multiplicity 2, which is even. Therefore, at (5/2, 0), the graph will touch the x-axis and turn around. The zero of -1 has multiplicity 1, which is odd. Therefore, at (-1, 0), the graph will cross through the x-axis.

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$$b)\hspace{.2em}$$ x-intercepts: $$(2, 0), (5, 0), \left(\frac{9}{2}, 0\right)$$ The zero of 2 has multiplicity 3, which is odd. Therefore, at (2, 0), the graph will cross through the x-axis. The zero of 5 has multiplicity 2, which is even. Therefore, at (5, 0), the graph will touch the x-axis and turn around. The zero of 9/2 has multiplicity 1, which is odd. Therefore, at (9/2, 0), the graph will cross through the x-axis.

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#4:

Solutions:

$$a)\hspace{.2em}$$ Since the degree of the polynomial function is 4, the maximum number of turning points is 3.

Local minimum points: $$(1, 2), (3, 2)$$ Local maximum points: $$\left(2, \frac{9}{4}\right)$$

$$b)\hspace{.2em}$$ Since the degree of the polynomial function is 3, the maximum number of turning points is 2.

Local minimum points: $$(1, 0)$$ Local maximum points: $$(3, 4)$$

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#5:

Solutions:

$$a)\hspace{.2em}$$

$$f(x) = 2(x - 1)(x - 5)(x - 3)$$ $$= 2x^3 - 18x^2 + 46x - 30$$ 1) The leading term is 2x^{3}. The leading coefficient is +2 and the degree of the leading term is 3, which is odd. The end behavior for the graph will be that it falls to the left and rises to the right.

2) The x-intercepts are (1, 0), (3, 0), and (5, 0). The multiplicity is 1 for each zero here, so the graph will cross through the x-axis at each x-intercept.

3) The y-intercept occurs at (0, -30). I'm going to limit the window to -10 to +10 for both the x-axis and y-axis, so this won't be visible on my graph. You may choose to change the scale of your graph, so it might be visible for you.

4) This function is not odd or even, so we won't use origin symmetry or y-axis symmetry.

Let's put what we have down on some graphing paper.

Now we can set up a table to get more points. When appropriate, we will round our values to the nearest hundredth.

Interval | x | y |
---|---|---|

(-∞, 1) | 0.75 | -4.78 |

(1, 3) | 2 | 6 |

(3, 5) | 4 | -6 |

(5, ∞) | 5.25 | 4.78 |

Let's update our graph to include the points from the table.

Just looking at the points on the graph, we can see there would be a turning point between an x-value of 1 and an x-value of 3.

Interval | x | y |
---|---|---|

(1, 3) | 1.5 | 5.25 |

(1, 3) | 1.65 | 5.88 |

(1, 3) | 1.75 | 6.09 |

(1, 3) | 1.8 | 6.14 |

(1, 3) | 1.85 | 6.16 |

(1, 3) | 1.9 | 6.14 |

(1, 3) | 1.95 | 6.08 |

(1, 3) | 2 | 6 |

(1, 3) | 2.25 | 5.16 |

(1, 3) | 2.5 | 3.75 |

Based on our table above, we would estimate our turn to be at an x-value of 1.85. Similarly, we know we have a turn between an x-value of 3 and an x-value of 5.

Interval | x | y |
---|---|---|

(3, 5) | 3.5 | -3.75 |

(3, 5) | 3.75 | -5.16 |

(3, 5) | 4 | -6 |

(3, 5) | 4.1 | -6.14 |

(3, 5) | 4.15 | -6.16 |

(3, 5) | 4.2 | -6.14 |

(3, 5) | 4.25 | -6.09 |

(3, 5) | 4.5 | -5.25 |

We see that the turn is going to happen at about an x-value of 4.15. Let's sketch our graph.

Desmos Link for More Detail$$b)\hspace{.2em}$$ $$f(x) = (2x - 3)^2(x - 4)^2$$ $$= 4x^4 - 44x^3 + 169x^2 - 264x + 144$$

1) The leading term is 4x^{4}. The leading coefficient is +4 and the degree of the leading term is 4, which is even. The end behavior for the graph will be that it rises to the left and rises to the right.

2) The x-intercepts are (3/2, 0) and (4, 0). The multiplicity is 2 for each zero here, so the graph will touch the x-axis at each x-intercept and turn around.

3) The y-intercept occurs at (0, 144). I'm going to limit the window to -5 to +15 for the y-axis and -10 to +10 for the x-axis, so this won't be visible on my graph. You may choose to change the scale of your graph, so it might be visible for you.

4) This function is not odd or even, so we won't use origin symmetry or y-axis symmetry.

Let's put what we have down on some graphing paper.

Now we can set up a table to get more points.

Interval | x | y |
---|---|---|

(-∞, 1.5) | 1 | 9 |

(1.5, 4) | 2 | 4 |

(1.5, 4) | 3 | 9 |

(4, ∞) | 4.5 | 9 |

Let's update our graph to include the points from the table.

Just looking at the points on the graph, we can see there would be a turning point between an x-value of 2 and an x-value of 4. We will round to the nearest hundredth where appropriate in our table below.

Interval | x | y |
---|---|---|

(2, 4) | 2.5 | 9 |

(2, 4) | 2.65 | 9.64 |

(2, 4) | 2.75 | 9.77 |

(2, 4) | 2.8 | 9.73 |

(2, 4) | 2.9 | 9.49 |

(2, 4) | 3 | 9 |

(2, 4) | 3.25 | 6.89 |

Based on our table above, we would estimate our turn to be at an x-value of 2.75. Let's sketch our graph.

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