Answer
**Summary of Solutions:**
1. **Asymptotes of \( f(x) \):**
- **Vertical Asymptote:** \( x = 3 \)
- **Horizontal Asymptote:** \( y = 2 \)
2. **Intercepts of \( f(x) \):**
- **\( x \)-Intercept:** \( (1, 0) \)
- **\( y \)-Intercept:** \( \left(0, \frac{2}{3}\right) \)
3. **Graphs of \( f(x) \) and \( g(x) \):**
- \( f(x) \) has a hyperbola shape with vertical asymptote at \( x = 3 \) and horizontal asymptote at \( y = 2 \).
- \( g(x) \) is a straight line with a slope of 1, crossing the \( y \)-axis at \( (0, 2) \) and the \( x \)-axis at \( (-2, 0) \).
4. **Points of Intersection:**
- \( (4, 6) \)
- \( (-1, 1) \)
5. **Inequality Solution:**
- For \( x < 3 \), \( \frac{4}{x - 3} + 2 < x + 2 \) holds true when \( -1 < x < 3 \).
---
**Note:** The graphs should be sketched on the same system of axes to visualize the intercepts, asymptotes, and points of intersection clearly.
Solution
Let's tackle each part of the problem step-by-step.
---
### **Given:**
- \( f(x) = \frac{4}{x - 3} + 2 \)
- \( g(x) = x + 2 \)
---
### **5.1. Equations of the Asymptotes of \( f \)**
**Vertical Asymptote:**
The vertical asymptote occurs where the denominator of the rational function is zero.
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
**Horizontal Asymptote:**
For rational functions of the form \( \frac{A}{x - h} + k \), as \( x \) approaches infinity or negative infinity, the \(\frac{A}{x - h}\) term approaches 0. Therefore, the horizontal asymptote is the constant term added to it.
\[
y = 2
\]
**Summary:**
- **Vertical Asymptote:** \( x = 3 \)
- **Horizontal Asymptote:** \( y = 2 \)
---
### **5.2. \( x \)-Intercept of \( f \)**
The \( x \)-intercept is found by setting \( f(x) = 0 \):
\[
\frac{4}{x - 3} + 2 = 0
\]
\[
\frac{4}{x - 3} = -2
\]
\[
4 = -2(x - 3)
\]
\[
4 = -2x + 6
\]
\[
-2 = -2x
\]
\[
x = 1
\]
**\( x \)-Intercept:** \( (1, 0) \)
---
### **5.3. \( y \)-Intercept of \( f \)**
The \( y \)-intercept is found by evaluating \( f(0) \):
\[
f(0) = \frac{4}{0 - 3} + 2 = \frac{4}{-3} + 2 = -\frac{4}{3} + 2 = -\frac{4}{3} + \frac{6}{3} = \frac{2}{3}
\]
**\( y \)-Intercept:** \( \left(0, \frac{2}{3}\right) \)
---
### **5.4. Sketching the Graphs of \( f \) and \( g \)**
**For \( f(x) = \frac{4}{x - 3} + 2 \):**
- **Vertical Asymptote:** \( x = 3 \)
- **Horizontal Asymptote:** \( y = 2 \)
- **\( x \)-Intercept:** \( (1, 0) \)
- **\( y \)-Intercept:** \( \left(0, \frac{2}{3}\right) \)
**For \( g(x) = x + 2 \):**
- **Slope:** 1
- **\( y \)-Intercept:** \( (0, 2) \)
- **\( x \)-Intercept:** Set \( g(x) = 0 \):
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
So, \( (-2, 0) \)
**Graph Description:**
- **\( f(x) \)** will have a hyperbola shape approaching the asymptotes \( x = 3 \) and \( y = 2 \), passing through \( (1, 0) \) and \( \left(0, \frac{2}{3}\right) \).
- **\( g(x) \)** is a straight line crossing the \( y \)-axis at \( (0, 2) \) and the \( x \)-axis at \( (-2, 0) \).
*Since this is a text medium, please plot the functions on graph paper or using graphing software to visualize their intersections, intercepts, and asymptotes clearly.*
---
### **5.5. \( x \)-Coordinates of Points of Intersection of \( f \) and \( g \)**
To find the points where \( f(x) = g(x) \):
\[
\frac{4}{x - 3} + 2 = x + 2
\]
Subtract 2 from both sides:
\[
\frac{4}{x - 3} = x
\]
Multiply both sides by \( (x - 3) \) (keeping in mind \( x \neq 3 \)):
\[
4 = x(x - 3)
\]
\[
4 = x^2 - 3x
\]
Bring all terms to one side:
\[
x^2 - 3x - 4 = 0
\]
Solve the quadratic equation:
\[
x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}
\]
**Solutions:**
- \( x = \frac{3 + 5}{2} = 4 \)
- \( x = \frac{3 - 5}{2} = -1 \)
**Points of Intersection:**
- \( (4, 6) \) since \( g(4) = 4 + 2 = 6 \)
- \( (-1, 1) \) since \( g(-1) = -1 + 2 = 1 \)
---
### **5.6. Values of \( x \) for Which \( \frac{4}{x - 3} + 2 < x + 2 \) When \( x < 3 \)**
We need to solve the inequality:
\[
\frac{4}{x - 3} + 2 < x + 2
\]
Subtract 2 from both sides:
\[
\frac{4}{x - 3} < x
\]
**Important:** Since \( x < 3 \), \( x - 3 < 0 \). When multiplying or dividing by a negative number, the inequality sign reverses.
Multiply both sides by \( (x - 3) \) (which is negative):
\[
4 > x(x - 3)
\]
\[
4 > x^2 - 3x
\]
Bring all terms to one side:
\[
x^2 - 3x - 4 < 0
\]
**Solve the Quadratic Inequality:**
First, find the roots of \( x^2 - 3x - 4 = 0 \):
\[
x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}
\]
\[
x = 4 \quad \text{and} \quad x = -1
\]
The quadratic \( x^2 - 3x - 4 \) is less than 0 between its roots:
\[
-1 < x < 4
\]
**But** since we're given \( x < 3 \), we intersect this with \( -1 < x < 4 \):
\[
-1 < x < 3
\]
**Conclusion:**
All real numbers \( x \) such that \( -1 < x < 3 \) satisfy the inequality \( \frac{4}{x - 3} + 2 < x + 2 \).
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Reviewed and approved by the UpStudy tutoring team
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