Answer :
Sure! Let's go through a detailed, step-by-step process to sketch the graph of the function [tex]\( f(x) = x^3 + 2x^2 - 8x \)[/tex]. We will include the analysis of critical points, concavity, zeros, and the intervals where the function is positive and negative.
### Step 1: Find the Zeros of the Function
The zeros of the function [tex]\( f(x) \)[/tex] are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].
Solving the equation [tex]\( x^3 + 2x^2 - 8x = 0 \)[/tex], we factor:
[tex]\[ x(x^2 + 2x - 8) = 0 \][/tex]
This gives us one zero directly:
[tex]\[ x = 0 \][/tex]
We then factor the quadratic:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) = 0 \][/tex]
This gives us additional zeros:
[tex]\[ x = -4 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the zeros of the function are: [tex]\( x = -4 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 2 \)[/tex].
### Step 2: Determine End-Behavior
To understand the end-behavior of [tex]\( f(x) \)[/tex], we examine what happens to [tex]\( f(x) \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex].
Since the leading term [tex]\( x^3 \)[/tex] dominates for large absolute values of [tex]\( x \)[/tex]:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
### Step 3: Find Critical Points and Test for Maxima/Minima
To find the critical points, we first compute the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 3x^2 + 4x - 8 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ 3x^2 + 4x - 8 = 0 \][/tex]
Solving this quadratic equation, we find:
[tex]\[ x = \frac{-2}{3} + \frac{2\sqrt{7}}{3} \][/tex]
[tex]\[ x = \frac{-2}{3} - \frac{2\sqrt{7}}{3} \][/tex]
### Step 4: Determine Concavity and Inflection Points
To determine concavity, we compute the second derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f''(x) = 6x + 4 \][/tex]
Setting the second derivative equal to zero to find the inflection points:
[tex]\[ 6x + 4 = 0 \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]
At [tex]\( x = -\frac{2}{3} \)[/tex], the concavity of the function changes, indicating an inflection point.
### Step 5: Determine Intervals of Positivity and Negativity
We analyze the intervals where the function is positive or negative.
- For [tex]\( (-\infty, -4) \)[/tex], the function is negative.
- For [tex]\( (-4, 0) \)[/tex], the function is positive.
- For [tex]\( (0, 2) \)[/tex], the function is negative.
- For [tex]\( (2, \infty) \)[/tex], the function is positive.
### Summary and Sketch
We now have all the information needed to sketch the function:
1. Zeros: [tex]\( x = -4 \)[/tex], [tex]\( x = 0 \)[/tex], [tex]\( x = 2 \)[/tex]
2. End-Behavior:
- [tex]\( f(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex]
- [tex]\( f(x) \to -\infty \)[/tex] as [tex]\( x \to -\infty \)[/tex]
3. Critical Points: [tex]\( x = \frac{-2}{3} + \frac{2\sqrt{7}}{3} \)[/tex], [tex]\( x = \frac{-2}{3} - \frac{2\sqrt{7}}{3} \)[/tex]
4. Inflection Point: [tex]\( x = -\frac{2}{3} \)[/tex]
5. Intervals:
- Negative: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (0, 2) \)[/tex]
- Positive: [tex]\( (-4, 0) \)[/tex], [tex]\( (2, \infty) \)[/tex]
By plotting these points and behavior, we can visually represent the function [tex]\( f(x) \)[/tex].
### Step 1: Find the Zeros of the Function
The zeros of the function [tex]\( f(x) \)[/tex] are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].
Solving the equation [tex]\( x^3 + 2x^2 - 8x = 0 \)[/tex], we factor:
[tex]\[ x(x^2 + 2x - 8) = 0 \][/tex]
This gives us one zero directly:
[tex]\[ x = 0 \][/tex]
We then factor the quadratic:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) = 0 \][/tex]
This gives us additional zeros:
[tex]\[ x = -4 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the zeros of the function are: [tex]\( x = -4 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 2 \)[/tex].
### Step 2: Determine End-Behavior
To understand the end-behavior of [tex]\( f(x) \)[/tex], we examine what happens to [tex]\( f(x) \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex].
Since the leading term [tex]\( x^3 \)[/tex] dominates for large absolute values of [tex]\( x \)[/tex]:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
### Step 3: Find Critical Points and Test for Maxima/Minima
To find the critical points, we first compute the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 3x^2 + 4x - 8 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ 3x^2 + 4x - 8 = 0 \][/tex]
Solving this quadratic equation, we find:
[tex]\[ x = \frac{-2}{3} + \frac{2\sqrt{7}}{3} \][/tex]
[tex]\[ x = \frac{-2}{3} - \frac{2\sqrt{7}}{3} \][/tex]
### Step 4: Determine Concavity and Inflection Points
To determine concavity, we compute the second derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f''(x) = 6x + 4 \][/tex]
Setting the second derivative equal to zero to find the inflection points:
[tex]\[ 6x + 4 = 0 \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]
At [tex]\( x = -\frac{2}{3} \)[/tex], the concavity of the function changes, indicating an inflection point.
### Step 5: Determine Intervals of Positivity and Negativity
We analyze the intervals where the function is positive or negative.
- For [tex]\( (-\infty, -4) \)[/tex], the function is negative.
- For [tex]\( (-4, 0) \)[/tex], the function is positive.
- For [tex]\( (0, 2) \)[/tex], the function is negative.
- For [tex]\( (2, \infty) \)[/tex], the function is positive.
### Summary and Sketch
We now have all the information needed to sketch the function:
1. Zeros: [tex]\( x = -4 \)[/tex], [tex]\( x = 0 \)[/tex], [tex]\( x = 2 \)[/tex]
2. End-Behavior:
- [tex]\( f(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex]
- [tex]\( f(x) \to -\infty \)[/tex] as [tex]\( x \to -\infty \)[/tex]
3. Critical Points: [tex]\( x = \frac{-2}{3} + \frac{2\sqrt{7}}{3} \)[/tex], [tex]\( x = \frac{-2}{3} - \frac{2\sqrt{7}}{3} \)[/tex]
4. Inflection Point: [tex]\( x = -\frac{2}{3} \)[/tex]
5. Intervals:
- Negative: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (0, 2) \)[/tex]
- Positive: [tex]\( (-4, 0) \)[/tex], [tex]\( (2, \infty) \)[/tex]
By plotting these points and behavior, we can visually represent the function [tex]\( f(x) \)[/tex].
