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Chapter 3: Problem 19

Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. $$P(x)=\frac{1}{12}(x+2)^{2}(x-3)^{2}$$

Short Answer

Expert verified
The graph is a degree 4 polynomial opening upwards, with x-intercepts at -2 and 3, and y-intercept at 3.

Step by step solution

01

Understand the Polynomial Function

The function given is \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), which is a polynomial of degree 4 because each factor \((x+2)\) and \((x-3)\) is squared. This indicates that the polynomial is of even degree, which gives it a particular end behavior.
02

Identify the Zeros and Multiplicity

The zeros of the function are derived from the factors: \(x+2=0\) and \(x-3=0\). Thus, the zeros are \(x = -2\) and \(x = 3\). Each zero has a multiplicity of 2 because each factor is squared, indicating that the graph will touch the x-axis at these points but not cross it.
03

Determine the End Behavior

Since the polynomial is of even degree (4) and the leading coefficient \(\frac{1}{12}\) is positive, the end behavior of the function is such that as \(x\) approaches both positive and negative infinity, \(P(x)\) approaches positive infinity. This suggests that both ends of the graph rise upwards.
04

Find the y-Intercept

The y-intercept is determined by evaluating \(P(x)\) at \(x=0\). \[P(0) = \frac{1}{12}(0+2)^2(0-3)^2 = \frac{1}{12}(4)(9) = 3.\] Thus, the y-intercept is \((0, 3)\).
05

Sketch the Graph

Begin by plotting the zeros \(x = -2\) and \(x = 3\), each as a point where the graph touches but does not cross the x-axis. Plot \((0, 3)\) as the y-intercept. The graph would start by rising in the second quadrant, touching at \(x = -2\), peaking on the positive y-axis, touching at \(x = 3\), and rising in the first quadrant, reflecting the end behavior.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intercepts of Polynomial Functions
When graphing polynomial functions, identifying the intercepts is crucial as they serve as key points on the graph. The intercepts are locations where the polynomial intersects the coordinate axes:
  • **x-intercepts (Roots/Zeros):** These are the points where the graph touches or crosses the x-axis. For a function like \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), the polynomial's factors tell us the zeros. Setting \( x+2 = 0 \) and \( x-3 = 0 \) gives the zeros \( x = -2 \) and \( x = 3 \), respectively. Each intercept is a crucial anchor for your graph.

  • **y-intercept:** This is where the graph intersects the y-axis. To find it, substitute \( x = 0 \) into the function: \( P(0) = \frac{1}{12}(4)(9) = 3 \). Thus, the y-intercept is at \( (0, 3) \).

Together, these intercepts help form a skeletal guide for plotting the polynomial's path.
End Behavior of Polynomials
Understanding the end behavior of a polynomial is like setting the boundaries for a road map. It allows us to see how the function behaves as it extends infinitely in either direction along the x-axis. Here are a few points to consider:
  • **Degree and Leading Coefficient:** For a polynomial like \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), the degree is 4 (sum of the exponents), which is even, and the leading coefficient is positive (\( \frac{1}{12} \)).

  • **Behavior Rule:** For even-degree polynomials with positive leading coefficients, such as in our example, as \( x \to \pm\infty \), the polynomial \( P(x) \to \infty \). Hence, both ends of the graph rise upwards, displaying a characteristic 'U' or 'W' shape depending on complexity. This is because both sides of the graph head upwards towards positive infinity.

Recognizing this pattern helps us predict the sketch without needing a point-by-point plot.
Zeros and Multiplicity
Zeros of a polynomial, also known as roots, indicate where the function equals zero. These points play a key role as they suggest where the graph touches or crosses the x-axis. Multiplicity further refines this behavior:
  • **Definition:** Multiplicity refers to the number of times a particular root is repeated. In \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), both zeros \( x = -2 \) and \( x = 3 \) have a multiplicity of 2, because each factor appears squared.

  • **Graphical Implication:** When a root's multiplicity is even, the graph touches the x-axis at the zero but doesn't cross it. This results in the graph curving back up or down, making a gentle U or upside-down U shape at these points. If the multiplicity is odd, it typically crosses the axis.

  • **Example Outcome:** In our specific function, \( P(x) \), the zeros will just touch the x-axis at \( x = -2 \) and \( x = 3 \) without crossing, as the multiplicities are 2. Understanding this ensures precision in sketching the outlines of the polynomial.
By knowing both zeros and their multiplicities, you can better visualise the turn and twist of the polynomial's graph as it navigates through these key points.

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Most popular questions from this chapter

Find the intercepts and asymptotes, and then sketch a graph of the rational function. Use a graphing device to confirm your answer. $$r(x)=\frac{(x-1)(x+2)}{(x+1)(x-3)}$$

The quadratic formula can be used to solve any quadratic (or second-degree) equation. You may have wondered if similar formulas exist for cubic (third- degree), quartic (fourth-degree), and higher-degree equations. For the depressed cubic \(x^{3}+p x+q=0\) Cardano (page 296 ) found the following formula for one solution: $$x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$ A formula for quartic equations was discovered by the Italian mathematician Ferrari in \(1540 .\) In 1824 the Norwegian mathematician Niels Henrik Abel proved that it is impossible to write a quintic formula, that is, a formula for fifth-degree equations. Finally, Galois (page 273 ) gave a criterion for determining which equations can be solved by a formula involving radicals. Use the cubic formula to find a solution for the following equations. Then solve the equations using the methods you learned in this section. Which method is easier? (a) \(x^{3}-3 x+2=0\) (b) \(x^{3}-27 x-54=0\) (c) \(x^{3}+3 x+4=0\)

Find the intercepts and asymptotes, and then sketch a graph of the rational function. Use a graphing device to confirm your answer. $$r(x)=\frac{2 x(x+2)}{(x-1)(x-4)}$$

Graph the rational function and find all vertical asymptotes, \(x\)- and \(y\)-intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$y=\frac{x^{4}-3 x^{3}+x^{2}-3 x+3}{x^{2}-3 x}$$

A polynomial \(P\) is given. (a) Find all the real zeros of \(P\). (b) Sketch the graph of \(P\). $$P(x)=x^{5}-x^{4}-5 x^{3}+x^{2}+8 x+4$$
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