Answer :
Certainly! Let's go through the steps needed to sketch the graph of the function [tex]\( f(x) = x^3 + 2x^2 - 8x \)[/tex]. We will include the end-behavior, zeroes, and intervals where the function is positive and negative.
### 1. Determine the end-behavior
To determine the end-behavior of the function, we look at the leading term, which is [tex]\( x^3 \)[/tex].
- As [tex]\( x \)[/tex] approaches infinity ([tex]\( x \to \infty \)[/tex]), [tex]\( x^3 \)[/tex] dominates and [tex]\( f(x) \)[/tex] approaches infinity: [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( x \to -\infty \)[/tex]), [tex]\( x^3 \)[/tex] also dominates but with a negative sign, so [tex]\( f(x) \to -\infty \)[/tex].
Thus, we can summarize the end-behavior as follows:
- As [tex]\( x \)[/tex] approaches infinity, [tex]\( f(x) \)[/tex] approaches infinity.
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] approaches negative infinity.
### 2. Find the zeroes of the function
The zeroes (roots) of the function are the points where [tex]\( f(x) = 0 \)[/tex].
Solving [tex]\( x^3 + 2x^2 - 8x = 0 \)[/tex]:
We can factor out the common term, [tex]\( x \)[/tex]:
[tex]\[ x(x^2 + 2x - 8) = 0 \][/tex]
This gives us one zero directly:
[tex]\[ x = 0 \][/tex]
Next, we factor the quadratic [tex]\( x^2 + 2x - 8 \)[/tex]:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) \][/tex]
So, the additional zeroes are:
[tex]\[ x = -4 \quad \text{and} \quad x = 2 \][/tex]
Therefore, the zeroes of [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -4, \quad x = 0, \quad \text{and} \quad x = 2 \][/tex]
### 3. Determine the critical points
The critical points are found where the derivative [tex]\( f'(x) \)[/tex] is zero or undefined.
First, compute the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 3x^2 + 4x - 8 \][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[ 3x^2 + 4x - 8 = 0 \][/tex]
Solving this quadratic equation, we find the critical points to be approximately:
[tex]\[ x = -2.431 \quad \text{and} \quad x = 1.097 \][/tex]
### 4. Determine where the function is positive or negative
Now, we need to determine the intervals where [tex]\( f(x) \)[/tex] is positive or negative. We use the zeroes and critical points to test the intervals:
The critical points divide the x-axis into three intervals:
- [tex]\( (-\infty, -2.431) \)[/tex]
- [tex]\( (-2.431, 1.097) \)[/tex]
- [tex]\( (1.097, \infty) \)[/tex]
Let's test a point in each interval to determine the sign of [tex]\( f(x) \)[/tex]:
- Interval [tex]\( (-\infty, -2.431) \)[/tex]:
- Choose [tex]\( x = -3 \)[/tex]
- [tex]\( f(-3) = (-3)^3 + 2(-3)^2 - 8(-3) = -27 + 18 + 24 > 0 \)[/tex] (Positive)
- Interval [tex]\( (-2.431, 1.097) \)[/tex]:
- Choose [tex]\( x = 0 \)[/tex]
- [tex]\( f(0) = 0 \)[/tex] (Positive)
- Interval [tex]\( (1.097, \infty) \)[/tex]:
- Choose [tex]\( x = 2 \)[/tex]
- [tex]\( f(2) = (2)^3 + 2(2)^2 - 8(2) = 8 + 8 - 16 = 0 \)[/tex] (Positive)
Therefore, the intervals where the function is positive and negative are:
- [tex]\( f(x) > 0 \)[/tex] on [tex]\( (-\infty, -2.431) \)[/tex]
- [tex]\( f(x) > 0 \)[/tex] on [tex]\( (-2.431, 1.097) \)[/tex]
- [tex]\( f(x) > 0 \)[/tex] on [tex]\( (1.097, \infty) \)[/tex]
### Summary of Key Points for the Graph
- End-behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
- Zeroes:
- The function has zeroes at [tex]\( x = -4, 0, \)[/tex] and [tex]\( 2 \)[/tex].
- Critical Points:
- The critical points are approximately at [tex]\( x = -2.431 \)[/tex] and [tex]\( x = 1.097 \)[/tex].
- Intervals:
- Positive on [tex]\( (-\infty, -2.431) \)[/tex]
- Positive on [tex]\( (-2.431, 1.097) \)[/tex]
- Positive on [tex]\( (1.097, \infty) \)[/tex]
Using this information, we can now sketch the graph of [tex]\( f(x) = x^3 + 2x^2 - 8x \)[/tex]. Note where the function crosses the x-axis, the directional behavior at the critical points, and the end-behavior as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] and [tex]\( -\infty \)[/tex].
### 1. Determine the end-behavior
To determine the end-behavior of the function, we look at the leading term, which is [tex]\( x^3 \)[/tex].
- As [tex]\( x \)[/tex] approaches infinity ([tex]\( x \to \infty \)[/tex]), [tex]\( x^3 \)[/tex] dominates and [tex]\( f(x) \)[/tex] approaches infinity: [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( x \to -\infty \)[/tex]), [tex]\( x^3 \)[/tex] also dominates but with a negative sign, so [tex]\( f(x) \to -\infty \)[/tex].
Thus, we can summarize the end-behavior as follows:
- As [tex]\( x \)[/tex] approaches infinity, [tex]\( f(x) \)[/tex] approaches infinity.
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] approaches negative infinity.
### 2. Find the zeroes of the function
The zeroes (roots) of the function are the points where [tex]\( f(x) = 0 \)[/tex].
Solving [tex]\( x^3 + 2x^2 - 8x = 0 \)[/tex]:
We can factor out the common term, [tex]\( x \)[/tex]:
[tex]\[ x(x^2 + 2x - 8) = 0 \][/tex]
This gives us one zero directly:
[tex]\[ x = 0 \][/tex]
Next, we factor the quadratic [tex]\( x^2 + 2x - 8 \)[/tex]:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) \][/tex]
So, the additional zeroes are:
[tex]\[ x = -4 \quad \text{and} \quad x = 2 \][/tex]
Therefore, the zeroes of [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -4, \quad x = 0, \quad \text{and} \quad x = 2 \][/tex]
### 3. Determine the critical points
The critical points are found where the derivative [tex]\( f'(x) \)[/tex] is zero or undefined.
First, compute the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 3x^2 + 4x - 8 \][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[ 3x^2 + 4x - 8 = 0 \][/tex]
Solving this quadratic equation, we find the critical points to be approximately:
[tex]\[ x = -2.431 \quad \text{and} \quad x = 1.097 \][/tex]
### 4. Determine where the function is positive or negative
Now, we need to determine the intervals where [tex]\( f(x) \)[/tex] is positive or negative. We use the zeroes and critical points to test the intervals:
The critical points divide the x-axis into three intervals:
- [tex]\( (-\infty, -2.431) \)[/tex]
- [tex]\( (-2.431, 1.097) \)[/tex]
- [tex]\( (1.097, \infty) \)[/tex]
Let's test a point in each interval to determine the sign of [tex]\( f(x) \)[/tex]:
- Interval [tex]\( (-\infty, -2.431) \)[/tex]:
- Choose [tex]\( x = -3 \)[/tex]
- [tex]\( f(-3) = (-3)^3 + 2(-3)^2 - 8(-3) = -27 + 18 + 24 > 0 \)[/tex] (Positive)
- Interval [tex]\( (-2.431, 1.097) \)[/tex]:
- Choose [tex]\( x = 0 \)[/tex]
- [tex]\( f(0) = 0 \)[/tex] (Positive)
- Interval [tex]\( (1.097, \infty) \)[/tex]:
- Choose [tex]\( x = 2 \)[/tex]
- [tex]\( f(2) = (2)^3 + 2(2)^2 - 8(2) = 8 + 8 - 16 = 0 \)[/tex] (Positive)
Therefore, the intervals where the function is positive and negative are:
- [tex]\( f(x) > 0 \)[/tex] on [tex]\( (-\infty, -2.431) \)[/tex]
- [tex]\( f(x) > 0 \)[/tex] on [tex]\( (-2.431, 1.097) \)[/tex]
- [tex]\( f(x) > 0 \)[/tex] on [tex]\( (1.097, \infty) \)[/tex]
### Summary of Key Points for the Graph
- End-behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
- Zeroes:
- The function has zeroes at [tex]\( x = -4, 0, \)[/tex] and [tex]\( 2 \)[/tex].
- Critical Points:
- The critical points are approximately at [tex]\( x = -2.431 \)[/tex] and [tex]\( x = 1.097 \)[/tex].
- Intervals:
- Positive on [tex]\( (-\infty, -2.431) \)[/tex]
- Positive on [tex]\( (-2.431, 1.097) \)[/tex]
- Positive on [tex]\( (1.097, \infty) \)[/tex]
Using this information, we can now sketch the graph of [tex]\( f(x) = x^3 + 2x^2 - 8x \)[/tex]. Note where the function crosses the x-axis, the directional behavior at the critical points, and the end-behavior as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] and [tex]\( -\infty \)[/tex].
