Answer :
Let's break down how to solve these polynomial function problems step by step. For each function, we'll do the following:
(a) Apply the Leading-Term Test: This involves looking at the term with the highest degree in the polynomial to understand the end behavior of the graph.
(b) Determine the Zeros and State the Multiplicity: Zeros are solutions to the equation when the function is set to zero. The multiplicity refers to the number of times a particular zero is repeated.
(c) Find a Few Additional Points: We’ll substitute a few values of [tex]\( x \)[/tex] into the function to see what [tex]\( y \)[/tex] values we get, which helps in sketching the graph.
(d) Graph the Function: We will use the information gathered to form a rough sketch of the graph.
Let's apply these steps to function 33: [tex]\( f(x) = x(x+4)(x-1)^2 \)[/tex].
Function 33:
(a) Leading-Term Test:
The leading term stems from multiplying the highest power from each factor:
- [tex]\( x \cdot x \cdot x^2 = x^4 \)[/tex]
So the leading term is [tex]\( x^4 \)[/tex]. Since the coefficient of [tex]\( x^4 \)[/tex] is positive, as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex] and as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
(b) Zeros and Multiplicity:
1. [tex]\( x = 0 \)[/tex] with multiplicity 1 (from [tex]\( x \)[/tex]).
2. [tex]\( x = -4 \)[/tex] with multiplicity 1 (from [tex]\( x + 4 \)[/tex]).
3. [tex]\( x = 1 \)[/tex] with multiplicity 2 (from [tex]\( (x - 1)^2 \)[/tex]).
(c) Additional Points:
Choose values for [tex]\( x \)[/tex] to find corresponding [tex]\( y \)[/tex] values:
- [tex]\( f(-2) = (-2)((-2)+4)((-2)-1)^2 = (-2)(2)(9) = -36 \)[/tex]
- [tex]\( f(0) = (0)(0+4)(0-1)^2 = 0 \)[/tex]
- [tex]\( f(2) = (2)((2)+4)((2)-1)^2 = (2)(6)(1) = 12 \)[/tex]
(d) Graph:
Using the information:
- The graph starts high and ends high (since the leading coefficient is positive and it’s an even degree).
- There are x-intercepts at [tex]\( x = -4 \)[/tex], [tex]\( x = 0 \)[/tex], and a bounce at [tex]\( x = 1 \)[/tex].
- Plot additional points to help shape the graph.
For the rest of the functions, you would follow a similar approach using these steps:
34. Function: [tex]\( f(x) = x^2(x-4)(x+2) \)[/tex]
- Leading term: [tex]\( x^4 \)[/tex] (positive coefficient indicates both ends of the graph rise).
- Zeros: [tex]\( x = 0 \)[/tex] (multiplicity 2), [tex]\( x = 4 \)[/tex] (multiplicity 1), [tex]\( x = -2 \)[/tex] (multiplicity 1).
- Additional points: You would compute similarly as above.
35. Function: [tex]\( f(x) = -x(x+3)^2(x-5) \)[/tex]
- Leading term: [tex]\(-x^4\)[/tex] (negative coefficient indicates the graph falls on both ends).
- Zeros: [tex]\( x = 0 \)[/tex] (multiplicity 1), [tex]\( x = -3 \)[/tex] (multiplicity 2), [tex]\( x = 5 \)[/tex] (multiplicity 1).
And similarly for the rest, focus on identifying the leading term, determining zeros and their multiplicities, finding additional points, and sketching the graph based on these findings. If you have any specific function you'd like further detailed assistance with, feel free to let me know!
(a) Apply the Leading-Term Test: This involves looking at the term with the highest degree in the polynomial to understand the end behavior of the graph.
(b) Determine the Zeros and State the Multiplicity: Zeros are solutions to the equation when the function is set to zero. The multiplicity refers to the number of times a particular zero is repeated.
(c) Find a Few Additional Points: We’ll substitute a few values of [tex]\( x \)[/tex] into the function to see what [tex]\( y \)[/tex] values we get, which helps in sketching the graph.
(d) Graph the Function: We will use the information gathered to form a rough sketch of the graph.
Let's apply these steps to function 33: [tex]\( f(x) = x(x+4)(x-1)^2 \)[/tex].
Function 33:
(a) Leading-Term Test:
The leading term stems from multiplying the highest power from each factor:
- [tex]\( x \cdot x \cdot x^2 = x^4 \)[/tex]
So the leading term is [tex]\( x^4 \)[/tex]. Since the coefficient of [tex]\( x^4 \)[/tex] is positive, as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex] and as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
(b) Zeros and Multiplicity:
1. [tex]\( x = 0 \)[/tex] with multiplicity 1 (from [tex]\( x \)[/tex]).
2. [tex]\( x = -4 \)[/tex] with multiplicity 1 (from [tex]\( x + 4 \)[/tex]).
3. [tex]\( x = 1 \)[/tex] with multiplicity 2 (from [tex]\( (x - 1)^2 \)[/tex]).
(c) Additional Points:
Choose values for [tex]\( x \)[/tex] to find corresponding [tex]\( y \)[/tex] values:
- [tex]\( f(-2) = (-2)((-2)+4)((-2)-1)^2 = (-2)(2)(9) = -36 \)[/tex]
- [tex]\( f(0) = (0)(0+4)(0-1)^2 = 0 \)[/tex]
- [tex]\( f(2) = (2)((2)+4)((2)-1)^2 = (2)(6)(1) = 12 \)[/tex]
(d) Graph:
Using the information:
- The graph starts high and ends high (since the leading coefficient is positive and it’s an even degree).
- There are x-intercepts at [tex]\( x = -4 \)[/tex], [tex]\( x = 0 \)[/tex], and a bounce at [tex]\( x = 1 \)[/tex].
- Plot additional points to help shape the graph.
For the rest of the functions, you would follow a similar approach using these steps:
34. Function: [tex]\( f(x) = x^2(x-4)(x+2) \)[/tex]
- Leading term: [tex]\( x^4 \)[/tex] (positive coefficient indicates both ends of the graph rise).
- Zeros: [tex]\( x = 0 \)[/tex] (multiplicity 2), [tex]\( x = 4 \)[/tex] (multiplicity 1), [tex]\( x = -2 \)[/tex] (multiplicity 1).
- Additional points: You would compute similarly as above.
35. Function: [tex]\( f(x) = -x(x+3)^2(x-5) \)[/tex]
- Leading term: [tex]\(-x^4\)[/tex] (negative coefficient indicates the graph falls on both ends).
- Zeros: [tex]\( x = 0 \)[/tex] (multiplicity 1), [tex]\( x = -3 \)[/tex] (multiplicity 2), [tex]\( x = 5 \)[/tex] (multiplicity 1).
And similarly for the rest, focus on identifying the leading term, determining zeros and their multiplicities, finding additional points, and sketching the graph based on these findings. If you have any specific function you'd like further detailed assistance with, feel free to let me know!
