Answer :
To sketch the function [tex]\( m(x) = x^4 + 2x^3 - 8x^2 \)[/tex], let's explore the behavior of the function across the interval from [tex]\( x = -10 \)[/tex] to [tex]\( x = 10 \)[/tex]. We'll consider key aspects such as the function's values at various points, its critical points, and its general shape.
### 1. Understand Basic Shape and Intercepts
First, recognize that the function [tex]\( m(x) = x^4 + 2x^3 - 8x^2 \)[/tex] is a polynomial of degree four, which gives us some insight into the general shape:
- Degree and Leading Coefficient: Since the highest power is [tex]\( x^4 \)[/tex] and its coefficient is positive, the ends of the graph will both rise to infinity as [tex]\( x \)[/tex] goes towards positive and negative infinity.
- Intercepts: The intercepts of the function can be found by setting [tex]\( m(x) = 0 \)[/tex]:
- [tex]\( x = 0 \)[/tex]: [tex]\( m(x) = 0^4 + 2 \cdot 0^3 - 8 \cdot 0^2 = 0 \)[/tex]
- Factorization or further algebra can be used to find other roots, but for sketching, knowing the behavior at [tex]\( x = 0 \)[/tex] suffices for now.
### 2. Critical Points and Stationary Points
Find the first derivative [tex]\( m'(x) \)[/tex]:
[tex]\[ m'(x) = 4x^3 + 6x^2 - 16x \][/tex]
Set [tex]\( m'(x) = 0 \)[/tex] to find critical points:
[tex]\[ 4x^3 + 6x^2 - 16x = 0 \][/tex]
Factor out the common term:
[tex]\[ 2x(2x^2 + 3x - 8) = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \][/tex]
[tex]\[ 2x^2 + 3x - 8 = 0 \][/tex]
The quadratic formula gives the roots of [tex]\( 2x^2 + 3x - 8 = 0 \)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-8)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 64}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{73}}{4} \][/tex]
Thus:
[tex]\[ x \approx 1.175 \quad \text{and} \quad x \approx -3.375 \][/tex]
### 3. Calculate and Plot Key Points
Using several points in the interval from [tex]\( x = -10 \)[/tex] to [tex]\( x = 10 \)[/tex] can help in sketching:
- At [tex]\( x = -10 \)[/tex], [tex]\( m(x) \approx 7200 \)[/tex]
- At [tex]\( x = -6 \)[/tex], [tex]\( m(x) \approx 873 \)[/tex]
- At [tex]\( x = -2 \)[/tex], [tex]\( m(x) \approx -9.71 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( m(x) = 0 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( m(x) \approx -6.33 \)[/tex]
- At [tex]\( x = 6 \)[/tex], [tex]\( m(x) \approx 899.33 \)[/tex]
- At [tex]\( x = 10 \)[/tex], [tex]\( m(x) \approx 12000 \)[/tex]
### 4. Sketch the Function
Plot the given points and connect them smoothly considering the polynomial behavior:
- The function has intercepts at [tex]\( x = 0 \)[/tex].
- Critical points around [tex]\( x \approx -3.375, 0, 1.175 \)[/tex] indicate local maxima and minima.
- The function rises towards positive infinity as [tex]\( x \)[/tex] approaches both positive and negative infinity.
```
Rough Sketch:
|m(x)_______|
7200 _____|
-10 | ___
| |
| |
| |
2999 | |
876 | |
| |
|
| |
|
|
0 -2-1-0-| --> [Local minimum near here]
|
|
|
-2 5-7| -------------------------
| _______________3
2999 | ----
11999 | ----
| - [xi]
y [ x=-10; x<1]
```
This sketch incorporates the general shape, showing important characteristics such as increasing and decreasing intervals, local maxima and minima, and end behavior.
### 1. Understand Basic Shape and Intercepts
First, recognize that the function [tex]\( m(x) = x^4 + 2x^3 - 8x^2 \)[/tex] is a polynomial of degree four, which gives us some insight into the general shape:
- Degree and Leading Coefficient: Since the highest power is [tex]\( x^4 \)[/tex] and its coefficient is positive, the ends of the graph will both rise to infinity as [tex]\( x \)[/tex] goes towards positive and negative infinity.
- Intercepts: The intercepts of the function can be found by setting [tex]\( m(x) = 0 \)[/tex]:
- [tex]\( x = 0 \)[/tex]: [tex]\( m(x) = 0^4 + 2 \cdot 0^3 - 8 \cdot 0^2 = 0 \)[/tex]
- Factorization or further algebra can be used to find other roots, but for sketching, knowing the behavior at [tex]\( x = 0 \)[/tex] suffices for now.
### 2. Critical Points and Stationary Points
Find the first derivative [tex]\( m'(x) \)[/tex]:
[tex]\[ m'(x) = 4x^3 + 6x^2 - 16x \][/tex]
Set [tex]\( m'(x) = 0 \)[/tex] to find critical points:
[tex]\[ 4x^3 + 6x^2 - 16x = 0 \][/tex]
Factor out the common term:
[tex]\[ 2x(2x^2 + 3x - 8) = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \][/tex]
[tex]\[ 2x^2 + 3x - 8 = 0 \][/tex]
The quadratic formula gives the roots of [tex]\( 2x^2 + 3x - 8 = 0 \)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-8)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 64}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{73}}{4} \][/tex]
Thus:
[tex]\[ x \approx 1.175 \quad \text{and} \quad x \approx -3.375 \][/tex]
### 3. Calculate and Plot Key Points
Using several points in the interval from [tex]\( x = -10 \)[/tex] to [tex]\( x = 10 \)[/tex] can help in sketching:
- At [tex]\( x = -10 \)[/tex], [tex]\( m(x) \approx 7200 \)[/tex]
- At [tex]\( x = -6 \)[/tex], [tex]\( m(x) \approx 873 \)[/tex]
- At [tex]\( x = -2 \)[/tex], [tex]\( m(x) \approx -9.71 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( m(x) = 0 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( m(x) \approx -6.33 \)[/tex]
- At [tex]\( x = 6 \)[/tex], [tex]\( m(x) \approx 899.33 \)[/tex]
- At [tex]\( x = 10 \)[/tex], [tex]\( m(x) \approx 12000 \)[/tex]
### 4. Sketch the Function
Plot the given points and connect them smoothly considering the polynomial behavior:
- The function has intercepts at [tex]\( x = 0 \)[/tex].
- Critical points around [tex]\( x \approx -3.375, 0, 1.175 \)[/tex] indicate local maxima and minima.
- The function rises towards positive infinity as [tex]\( x \)[/tex] approaches both positive and negative infinity.
```
Rough Sketch:
|m(x)_______|
7200 _____|
-10 | ___
| |
| |
| |
2999 | |
876 | |
| |
|
| |
|
|
0 -2-1-0-| --> [Local minimum near here]
|
|
|
-2 5-7| -------------------------
| _______________3
2999 | ----
11999 | ----
| - [xi]
y [ x=-10; x<1]
```
This sketch incorporates the general shape, showing important characteristics such as increasing and decreasing intervals, local maxima and minima, and end behavior.
