Problem 21
Question
Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. $$ P(x)=x^{3}(x+2)(x-3)^{2} $$
Step-by-Step Solution
Verified Answer
The graph crosses at \(x = -2\), flattens at \(x = 0\), and touches at \(x = 3\); both ends face upwards.
1Step 1: Identify the Degree and End Behavior
The polynomial is given as \( P(x) = x^3(x+2)(x-3)^2 \). Firstly, identify the degree of the polynomial by adding the exponents of all factors. Here, \( x^3 \) contributes 3, the factor \( (x+2) \) contributes 1, and \( (x-3)^2 \) contributes 2 to the degree. Hence, the total degree is \( 3+1+2 = 6 \). Since it is an even degree polynomial with a positive leading coefficient, the end behavior is upwards on both sides: as \( x \to \pm\infty \), \( P(x) \to +\infty \).
2Step 2: Find the Zeros of the Polynomial
The zeros of the polynomial can be found by setting each factor to zero. Solving \( x^3 = 0 \) gives \( x = 0 \) with multiplicity 3, solving \( x + 2 = 0 \) gives \( x = -2 \) with multiplicity 1, and \( (x - 3)^2 = 0 \) gives \( x = 3 \) with multiplicity 2.
3Step 3: Determine the Behavior Around Each Zero
Factor behaviors depend on their multiplicity. At vertex \( x = 0 \), the polynomial will flatten due to multiplicity 3. The zero at \( x = -2 \) (multiplicity 1) results in a cross, and at \( x = 3 \) (multiplicity 2), the curve will touch the x-axis and turn back.
4Step 4: Calculate the Y-Intercept
To find the y-intercept, substitute \( x = 0 \) into the polynomial. This results in \( P(0) = 0^3(0+2)(0-3)^2 = 0 \). The y-intercept is thus at the origin \( (0,0) \).
5Step 5: Sketch the Graph
Given the end behavior and the characteristics around each zero, draw the graph: Start from the left, rising as \( x \to +\infty \). Cross the x-axis at \( x = -2 \), flatten briefly at the origin, touch and turn away at \( x = 3 \), and continue rising towards the right. Mark important points such as \(( -2, 0) \), \((0, 0) \), and \(( 3, 0) \).
Key Concepts
End BehaviorPolynomial ZerosGraph InterceptsMultiplicity of Roots
End Behavior
The end behavior of a polynomial describes what happens to the values of the polynomial as the input variable, usually denoted as \( x \), approaches positive or negative infinity. For the polynomial \( P(x) = x^3(x+2)(x-3)^2 \), the degree of the polynomial is calculated by adding up all the exponents: 3 from \( x^3 \), 1 from \( x+2 \), and 2 from \( (x-3)^2 \), summing to a total degree of 6.
Since the degree of the polynomial is even (6), and the leading term coefficient (the coefficient of the highest degree term) is positive, the end behavior is characterized by both ends of the graph going upwards. This means as \( x \to \infty \), \( P(x) \to \infty \) and as \( x \to -\infty \), \( P(x) \to \infty \) as well. Understanding this helps to predict the shape of the graph at the extremes.
Since the degree of the polynomial is even (6), and the leading term coefficient (the coefficient of the highest degree term) is positive, the end behavior is characterized by both ends of the graph going upwards. This means as \( x \to \infty \), \( P(x) \to \infty \) and as \( x \to -\infty \), \( P(x) \to \infty \) as well. Understanding this helps to predict the shape of the graph at the extremes.
Polynomial Zeros
Finding the zeros of a polynomial involves solving each factor of the polynomial equation set to zero. For our polynomial \( P(x) = x^3(x+2)(x-3)^2 \), the zeros occur when each factor is individually set to zero.
- For \( x^3 = 0 \), the solution is \( x = 0 \) with a multiplicity of 3.
- For \( x+2 = 0 \), the solution is \( x = -2 \) with a multiplicity of 1.
- For \( (x-3)^2 = 0 \), the solution is \( x = 3 \) with a multiplicity of 2.
Graph Intercepts
Intercepts are points where the graph crosses the axes. For polynomials, the x-intercepts are the zeros, while the y-intercept is found by evaluating the polynomial at \( x = 0 \).
The x-intercepts for \( P(x) = x^3(x+2)(x-3)^2 \) are at the zero locations we calculated: \( x = -2 \), \( x = 0 \), and \( x = 3 \). These correspond to the points \((-2,0) \), \((0,0) \), and \((3,0) \) respectively.
To find the y-intercept, calculate \( P(0) \):
\[ P(0) = 0^3(0+2)(0-3)^2 = 0 \]
Thus, the y-intercept is at \((0,0) \), which is the origin. Knowing both x and y intercepts gives us crucial plotting points for the graph.
The x-intercepts for \( P(x) = x^3(x+2)(x-3)^2 \) are at the zero locations we calculated: \( x = -2 \), \( x = 0 \), and \( x = 3 \). These correspond to the points \((-2,0) \), \((0,0) \), and \((3,0) \) respectively.
To find the y-intercept, calculate \( P(0) \):
\[ P(0) = 0^3(0+2)(0-3)^2 = 0 \]
Thus, the y-intercept is at \((0,0) \), which is the origin. Knowing both x and y intercepts gives us crucial plotting points for the graph.
Multiplicity of Roots
The multiplicity of a root refers to the number of times a particular solution appears in the factorization of a polynomial. This concept affects how the graph behaves at the corresponding zero.
- A root with an odd multiplicity, like \( x = 0 \) with multiplicity 3, means the graph crosses the x-axis at that point, but it will `flatten` due to the multiplicity being higher than 1.
- A root with a multiplicity of 1, as found at \( x = -2 \), denotes a clean crossing of the x-axis, where the graph forms a typical linear intersection.
- At \( x = 3 \), since the root has a multiplicity of 2 (which is even), the graph will touch the x-axis and `bounce` back, creating a vertex but not crossing it.
Other exercises in this chapter
Problem 21
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View solution Problem 21
Find all rational zeros of the polynomial. $$ P(x)=x^{4}-5 x^{2}+4 $$
View solution Problem 22
13- 30 . Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. $$ P(x)=x^{3}-64 $$
View solution Problem 22
Find all horizontal and vertical asymptotes (if any). \(s(x)=\frac{3 x^{2}}{x^{2}+2 x+5}\)
View solution