Question
Question content area top left
Part 1
Answer parts (a)dash(e) for the function shown below.
f(x)equals x cubed plus 3 x squared minus x minus 3
Question content area bottom left
Part 1
a. Use the leading coefficient test to determine the graph's end behavior. Which statement describes the behavior at the ends of f(x)equals x cubed plus 3 x squared minus x minus 3?
A.
The graph falls to the left and rises to the right.
B.
The graph rises to the left and to the right.
C.
The graph rises to the left and falls to the right.
D.
The graph falls to the left and to the right.
.
.
.
Question content area right
Part 1
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b. Find the x-intercepts. State whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept. What are the x-intercepts?
xequals
enter your response here
(Type an integer or a decimal. Use a comma to separate answers as needed.)
At which x-intercept(s) does the graph cross the x-axis? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
xequals
enter your response here
(Type an integer or a decimal. Use a comma to separate answers as needed.)
B.
There are no x-intercepts at which the graph crosses the x-axis.
At which x-intercept(s) does the graph touch the x-axis and turn around? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
xequals
enter your response here
(Type an integer or a decimal. Use a comma to separate answers as needed.)
B.
There are no x-intercepts at which the graph touches the x-axis and turns around.
c. Find the y-intercept.
The y-intercept is yequals
enter your response here.
(Type an integer or a decimal.)
Part 1
Answer parts (a)dash(e) for the function shown below.
f(x)equals x cubed plus 3 x squared minus x minus 3
Question content area bottom left
Part 1
a. Use the leading coefficient test to determine the graph's end behavior. Which statement describes the behavior at the ends of f(x)equals x cubed plus 3 x squared minus x minus 3?
A.
The graph falls to the left and rises to the right.
B.
The graph rises to the left and to the right.
C.
The graph rises to the left and falls to the right.
D.
The graph falls to the left and to the right.
.
.
.
Question content area right
Part 1
-10
-8
-6
-4
-2
2
4
6
8
10
-10
-8
-6
-4
-2
2
4
6
8
10
x
y
b. Find the x-intercepts. State whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept. What are the x-intercepts?
xequals
enter your response here
(Type an integer or a decimal. Use a comma to separate answers as needed.)
At which x-intercept(s) does the graph cross the x-axis? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
xequals
enter your response here
(Type an integer or a decimal. Use a comma to separate answers as needed.)
B.
There are no x-intercepts at which the graph crosses the x-axis.
At which x-intercept(s) does the graph touch the x-axis and turn around? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
xequals
enter your response here
(Type an integer or a decimal. Use a comma to separate answers as needed.)
B.
There are no x-intercepts at which the graph touches the x-axis and turns around.
c. Find the y-intercept.
The y-intercept is yequals
enter your response here.
(Type an integer or a decimal.)
Answers
GPT-4o mini
Let's analyze the function \( f(x) = x^3 + 3x^2 - x - 3 \) step-by-step to answer parts (a), (b), and (c).
### Part (a): Leading Coefficient Test
The leading term of the polynomial \( f(x) \) is \( x^3 \). The leading coefficient is \( 1 \), which is positive.
Since the degree (3, which is odd) is also odd, we can apply the leading coefficient test:
- As \( x \to -\infty \), \( f(x) \to -\infty \) (the graph falls to the left).
- As \( x \to +\infty \), \( f(x) \to +\infty \) (the graph rises to the right).
Thus, the answer is:
**A. The graph falls to the left and rises to the right.**
---
### Part (b): Finding the \( x \)-intercepts
To find the \( x \)-intercepts, we set \( f(x) = 0 \):
\[
x^3 + 3x^2 - x - 3 = 0
\]
We can factor or find the roots of the polynomial. This polynomial can be factored using synthetic division or by trying some possible rational roots (like \( x = 1 \) or \( x = -1 \)).
After attempting \( x = 1 \):
\[
f(1) = 1^3 + 3(1)^2 - 1 - 3 = 1 + 3 - 1 - 3 = 0
\]
So, \( x = 1 \) is an \( x \)-intercept.
Now we perform synthetic division of \( f(x) \) by \( x - 1 \):
\[
\begin{array}{r|rrrr}
1 & 1 & 3 & -1 & -3 \\
& & 1 & 4 & 3 \\
\hline
& 1 & 4 & 3 & 0 \\
\end{array}
\]
This gives us the quotient \( x^2 + 4x + 3 \).
Now we can factor \( x^2 + 4x + 3 \):
\[
x^2 + 4x + 3 = (x + 1)(x + 3)
\]
Thus, the complete factorization of \( f(x) \) is:
\[
f(x) = (x - 1)(x + 1)(x + 3)
\]
The \( x \)-intercepts are:
\[
x = 1, \quad x = -1, \quad x = -3
\]
To determine whether the graph crosses or touches the \( x \)-axis:
- The intercepts \( x = 1 \), \( x = -1 \), and \( x = -3 \) all have odd multiplicity (all are first degree). This means the graph crosses the \( x \)-axis at all of these intercepts.
Thus we conclude:
**The \( x \)-intercepts are:
\( x = 1, -1, -3 \)**
For crossing and touching:
- **Crosses at:** \( x = 1, -1, -3 \).
- **Touches and turns around:** None.
### Part (c): Finding the \( y \)-intercept
To find the \( y \)-intercept, substitute \( x = 0 \) into the function:
\[
f(0) = 0^3 + 3(0)^2 - 0 - 3 = -3
\]
Thus, the \( y \)-intercept is:
**The \( y \)-intercept is \( y = -3 \)**.
---
To summarize:
- **(a)** A: The graph falls to the left and rises to the right.
- **(b)** \( x \)-intercepts: \( x = 1, -1, -3 \); crosses at all three intercepts.
- **(c)** \( y \)-intercept: \( y = -3 \).
### Part (a): Leading Coefficient Test
The leading term of the polynomial \( f(x) \) is \( x^3 \). The leading coefficient is \( 1 \), which is positive.
Since the degree (3, which is odd) is also odd, we can apply the leading coefficient test:
- As \( x \to -\infty \), \( f(x) \to -\infty \) (the graph falls to the left).
- As \( x \to +\infty \), \( f(x) \to +\infty \) (the graph rises to the right).
Thus, the answer is:
**A. The graph falls to the left and rises to the right.**
---
### Part (b): Finding the \( x \)-intercepts
To find the \( x \)-intercepts, we set \( f(x) = 0 \):
\[
x^3 + 3x^2 - x - 3 = 0
\]
We can factor or find the roots of the polynomial. This polynomial can be factored using synthetic division or by trying some possible rational roots (like \( x = 1 \) or \( x = -1 \)).
After attempting \( x = 1 \):
\[
f(1) = 1^3 + 3(1)^2 - 1 - 3 = 1 + 3 - 1 - 3 = 0
\]
So, \( x = 1 \) is an \( x \)-intercept.
Now we perform synthetic division of \( f(x) \) by \( x - 1 \):
\[
\begin{array}{r|rrrr}
1 & 1 & 3 & -1 & -3 \\
& & 1 & 4 & 3 \\
\hline
& 1 & 4 & 3 & 0 \\
\end{array}
\]
This gives us the quotient \( x^2 + 4x + 3 \).
Now we can factor \( x^2 + 4x + 3 \):
\[
x^2 + 4x + 3 = (x + 1)(x + 3)
\]
Thus, the complete factorization of \( f(x) \) is:
\[
f(x) = (x - 1)(x + 1)(x + 3)
\]
The \( x \)-intercepts are:
\[
x = 1, \quad x = -1, \quad x = -3
\]
To determine whether the graph crosses or touches the \( x \)-axis:
- The intercepts \( x = 1 \), \( x = -1 \), and \( x = -3 \) all have odd multiplicity (all are first degree). This means the graph crosses the \( x \)-axis at all of these intercepts.
Thus we conclude:
**The \( x \)-intercepts are:
\( x = 1, -1, -3 \)**
For crossing and touching:
- **Crosses at:** \( x = 1, -1, -3 \).
- **Touches and turns around:** None.
### Part (c): Finding the \( y \)-intercept
To find the \( y \)-intercept, substitute \( x = 0 \) into the function:
\[
f(0) = 0^3 + 3(0)^2 - 0 - 3 = -3
\]
Thus, the \( y \)-intercept is:
**The \( y \)-intercept is \( y = -3 \)**.
---
To summarize:
- **(a)** A: The graph falls to the left and rises to the right.
- **(b)** \( x \)-intercepts: \( x = 1, -1, -3 \); crosses at all three intercepts.
- **(c)** \( y \)-intercept: \( y = -3 \).