Answer :
It seems you have a set of diverse questions, which will require both graphical and algebraic solutions. Let's handle each one step-by-step.
### QUESTION 1
#### 1.1 Graph of [tex]\( y = -x^3 \)[/tex]
The function [tex]\( y = -x^3 \)[/tex] is a cubic function reflected over the x-axis. Sketching this, you would get an S-shaped curve starting from the lower left, passing through the origin, and then turning to the lower right. Here is a simplified version:
```
y
|
-3 +
-2 +
-1 +
0 +-----------------+---------
-3 -2 -1 0 1 2 3 x
1 -
2 -
3 -
```
#### 1.2 Graph of the Inverse of [tex]\( y = x^2 - 4 \)[/tex]
To find the inverse, we switch [tex]\( x \)[/tex] and [tex]\( y \)[/tex] and solve for [tex]\( y \)[/tex]:
[tex]\[ x = y^2 - 4 \][/tex]
[tex]\[ y^2 = x + 4 \][/tex]
[tex]\[ y = \pm \sqrt{x + 4} \][/tex]
This represents two functions: [tex]\( y = \sqrt{x + 4} \)[/tex] and [tex]\( y = -\sqrt{x + 4} \)[/tex]. Sketching both gives the graph of the inverse:
```
y
4 |
3 |
2 |
1 |
0 |--------+-------------+----- x
-1 |
-2 |
-3 |
-4 |
```
#### 1.3 Graph of [tex]\( xy = -4 \)[/tex]
This is a hyperbola. We can rewrite it as [tex]\( y = -\frac{4}{x} \)[/tex]. The hyperbola has two branches with asymptotes at y=0 and x=0.
```
y
3 |
2 |
1 |
0 +-----------------+------
-1 |
-2 |
-3 |
-3 -2 -1 0 1 2 3
```
#### 1.4 Graph of [tex]\( y = \pm \sqrt{16 - x^2} \)[/tex]
This describes the upper and lower halves of a circle centered at the origin with radius 4. Sketching this:
```
y
4 +
3 +
2 +
1 +
0 +-----------------+------ x
-1 +
-2 +
-3 +
-4 +
```
### QUESTION 2
#### 2.1 Simplify the following:
##### 2.1.1 [tex]\( (3 - 2j)^4 \)[/tex]
First, convert to polar form:
1. Compute the magnitude:
[tex]\[ r = |3 - 2j| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \][/tex]
2. Compute the argument:
[tex]\[ \theta = \tan^{-1}\left(\frac{-2}{3}\right) \][/tex]
3. Express in polar form:
[tex]\[ (3 - 2j) \rightarrow \sqrt{13} \text{cis} \left(\theta\right) \][/tex]
Raising to the power of 4:
[tex]\[ (\sqrt{13} \text{cis} \left(\theta\right))^4 = (\sqrt{13})^4 \text{cis}(4\theta) = 169 \text{cis}(4\theta) \][/tex]
##### 2.1.2
[tex]\[ \frac{(2 + j)(-4 + 2j)}{3 - 2j} \][/tex]
First, multiply the numerators:
[tex]\[ (2+j)(-4+2j) = -8 + 4j - 4j - 2j^2 = -8 + 2(-1) = -8 - 2 = -10 \][/tex]
Then, divide by [tex]\(3 - 2j\)[/tex]:
[tex]\[ \text{Use complex conjugate:} \\ \frac{-10}{3 - 2j} = \frac{-10(3 + 2j)}{(3 - 2j)(3 + 2j)} = \frac{-30 - 20j}{9 + 4} = \frac{-30 - 20j}{13} \][/tex]
Express as:
[tex]\[ -\frac{30}{13} - j\frac{20}{13} \][/tex]
##### 2.1.3
[tex]\[ \frac{3j^6 - j^{16}}{1 - 2j} \][/tex]
Calculate the powers of [tex]\(j\)[/tex]:
- [tex]\(j^6 = (j^2)^3 = (-1)^3 = -1\)[/tex]
- [tex]\(j^{16} = (j^4)^4 = 1^4 = 1\)[/tex]
So:
[tex]\[3j^6 - j^{16} = 3(-1) - 1 = -3 - 1 = -4\][/tex]
Now divide by [tex]\(1 - 2j\)[/tex] using the complex conjugate:
[tex]\[ -\frac{4}{1 - 2j} \cdot \frac{1 + 2j}{1 + 2j} = \frac{-4 - 8j}{1 + 4} = \frac{-4 - 8j}{5} = -\frac{4}{5} - j\frac{8}{5} \][/tex]
Expressed in the form [tex]\(a + bj\)[/tex]:
[tex]\[ -\frac{4}{5} - j\frac{8}{5} \][/tex]
### QUESTION 1
#### 1.1 Graph of [tex]\( y = -x^3 \)[/tex]
The function [tex]\( y = -x^3 \)[/tex] is a cubic function reflected over the x-axis. Sketching this, you would get an S-shaped curve starting from the lower left, passing through the origin, and then turning to the lower right. Here is a simplified version:
```
y
|
-3 +
-2 +
-1 +
0 +-----------------+---------
-3 -2 -1 0 1 2 3 x
1 -
2 -
3 -
```
#### 1.2 Graph of the Inverse of [tex]\( y = x^2 - 4 \)[/tex]
To find the inverse, we switch [tex]\( x \)[/tex] and [tex]\( y \)[/tex] and solve for [tex]\( y \)[/tex]:
[tex]\[ x = y^2 - 4 \][/tex]
[tex]\[ y^2 = x + 4 \][/tex]
[tex]\[ y = \pm \sqrt{x + 4} \][/tex]
This represents two functions: [tex]\( y = \sqrt{x + 4} \)[/tex] and [tex]\( y = -\sqrt{x + 4} \)[/tex]. Sketching both gives the graph of the inverse:
```
y
4 |
3 |
2 |
1 |
0 |--------+-------------+----- x
-1 |
-2 |
-3 |
-4 |
```
#### 1.3 Graph of [tex]\( xy = -4 \)[/tex]
This is a hyperbola. We can rewrite it as [tex]\( y = -\frac{4}{x} \)[/tex]. The hyperbola has two branches with asymptotes at y=0 and x=0.
```
y
3 |
2 |
1 |
0 +-----------------+------
-1 |
-2 |
-3 |
-3 -2 -1 0 1 2 3
```
#### 1.4 Graph of [tex]\( y = \pm \sqrt{16 - x^2} \)[/tex]
This describes the upper and lower halves of a circle centered at the origin with radius 4. Sketching this:
```
y
4 +
3 +
2 +
1 +
0 +-----------------+------ x
-1 +
-2 +
-3 +
-4 +
```
### QUESTION 2
#### 2.1 Simplify the following:
##### 2.1.1 [tex]\( (3 - 2j)^4 \)[/tex]
First, convert to polar form:
1. Compute the magnitude:
[tex]\[ r = |3 - 2j| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \][/tex]
2. Compute the argument:
[tex]\[ \theta = \tan^{-1}\left(\frac{-2}{3}\right) \][/tex]
3. Express in polar form:
[tex]\[ (3 - 2j) \rightarrow \sqrt{13} \text{cis} \left(\theta\right) \][/tex]
Raising to the power of 4:
[tex]\[ (\sqrt{13} \text{cis} \left(\theta\right))^4 = (\sqrt{13})^4 \text{cis}(4\theta) = 169 \text{cis}(4\theta) \][/tex]
##### 2.1.2
[tex]\[ \frac{(2 + j)(-4 + 2j)}{3 - 2j} \][/tex]
First, multiply the numerators:
[tex]\[ (2+j)(-4+2j) = -8 + 4j - 4j - 2j^2 = -8 + 2(-1) = -8 - 2 = -10 \][/tex]
Then, divide by [tex]\(3 - 2j\)[/tex]:
[tex]\[ \text{Use complex conjugate:} \\ \frac{-10}{3 - 2j} = \frac{-10(3 + 2j)}{(3 - 2j)(3 + 2j)} = \frac{-30 - 20j}{9 + 4} = \frac{-30 - 20j}{13} \][/tex]
Express as:
[tex]\[ -\frac{30}{13} - j\frac{20}{13} \][/tex]
##### 2.1.3
[tex]\[ \frac{3j^6 - j^{16}}{1 - 2j} \][/tex]
Calculate the powers of [tex]\(j\)[/tex]:
- [tex]\(j^6 = (j^2)^3 = (-1)^3 = -1\)[/tex]
- [tex]\(j^{16} = (j^4)^4 = 1^4 = 1\)[/tex]
So:
[tex]\[3j^6 - j^{16} = 3(-1) - 1 = -3 - 1 = -4\][/tex]
Now divide by [tex]\(1 - 2j\)[/tex] using the complex conjugate:
[tex]\[ -\frac{4}{1 - 2j} \cdot \frac{1 + 2j}{1 + 2j} = \frac{-4 - 8j}{1 + 4} = \frac{-4 - 8j}{5} = -\frac{4}{5} - j\frac{8}{5} \][/tex]
Expressed in the form [tex]\(a + bj\)[/tex]:
[tex]\[ -\frac{4}{5} - j\frac{8}{5} \][/tex]
