Question

Sketch the graph of the following rational functions and identified the intercept, y-intercept
If it exists and the vertical symptote and range of the function.
A. Y÷3+x÷12=4
B. 4y=x_2
C. Y=x²_6x_7
D. Y= 2x÷x_1
E. Y=x_1÷x²_9
F. Y= square root x+4

Answer 1

Let's analyze each function in detail to identify intercepts, vertical asymptotes, and ranges. We'll start by rewriting and solving each function step-by-step.

### A. \( \frac{y}{3} + \frac{x}{12} = 4 \)

1. Rearrange the equation:
\[
\frac{y}{3} = 4 - \frac{x}{12} \\
y = 12 - \frac{x}{4}
\]
This is a linear equation in slope-intercept form \( y = mx + b \).

2. **Intercepts**:
- **Y-intercept**: Set \( x = 0 \):
\[
y = 12 \implies (0, 12)
\]
- **X-intercept**: Set \( y = 0 \):
\[
0 = 12 - \frac{x}{4} \implies \frac{x}{4} = 12 \implies x = 48 \implies (48, 0)
\]

3. **Vertical Asymptote**: Not applicable since it's a linear function.

4. **Range**: Since it's a linear function, the range is all real numbers: \( (-\infty, \infty) \).

### B. \( 4y = x^2 \)

1. Rearranging gives:
\[
y = \frac{x^2}{4}
\]

2. **Intercepts**:
- **Y-intercept**: Set \( x = 0 \):
\[
y = 0 \implies (0, 0)
\]
- **X-intercept**: Set \( y = 0 \):
\[
0 = \frac{x^2}{4} \implies x = 0 \implies (0, 0)
\]

3. **Vertical Asymptote**: Not applicable since it's a quadratic function.

4. **Range**: The minimum value is \( 0 \), so the range is \( [0, \infty) \).

### C. \( y = x^2 - 6x + 7 \)

1. This is a quadratic function, and we can find its vertex and intercepts:
\[
y = (x - 3)^2 - 2
\]

2. **Intercepts**:
- **Y-intercept**: Set \( x = 0 \):
\[
y = 0^2 - 6(0) + 7 = 7 \implies (0, 7)
\]
- **X-intercepts**: Set \( y = 0 \):
\[
0 = x^2 - 6x + 7 \implies \text{Use the quadratic formula or complete the square.} \\
x = \frac{6 \pm \sqrt{36 - 28}}{2} = 3 \pm \sqrt{2} \implies (3+\sqrt{2}, 0) \text{ and } (3-\sqrt{2}, 0)
\]

3. **Vertical Asymptote**: Not applicable since it's a polynomial.

4. **Range**: The minimum value is \( -2 \) (the vertex), so the range is \( [-2, \infty) \).

### D. \( y = \frac{2x}{x - 1} \)

1. This is a rational function.

2. **Intercepts**:
- **Y-intercept**: Set \( x = 0 \):
\[
y = \frac{2(0)}{0 - 1} = 0 \implies (0, 0)
\]
- **X-intercept**: Set \( y = 0 \):
\[
0 = \frac{2x}{x - 1} \implies x = 0 \implies (0, 0)
\]

3. **Vertical Asymptote**: Set the denominator to zero:
\[
x - 1 = 0 \implies x = 1
\]

4. **Range**: The horizontal asymptote \( y = 2 \). Thus, the range is \( (-\infty, 2) \cup (2, \infty) \).

### E. \( y = \frac{1}{x^2 - 9} \)

1. This is also a rational function.

2. **Intercepts**:
- **Y-intercept**: Set \( x = 0 \):
\[
y = \frac{1}{0 - 9} = -\frac{1}{9} \implies (0, -\frac{1}{9})
\]
- **X-intercept**: There is none since the numerator is constant and non-zero.

3. **Vertical Asymptotes**: Set the denominator to zero:
\[
x^2 - 9 = 0 \implies x = 3 \text{ and } x = -3
\]

4. **Range**: The function approaches 0 but never reaches it; therefore, the range is \( (-\infty, 0) \).

### F. \( y = \sqrt{x + 4} \)

1. This is a square root function.

2. **Intercepts**:
- **Y-intercept**: Set \( x = 0 \):
\[
y = \sqrt{0 + 4} = 2 \implies (0, 2)
\]
- **X-intercept**: Set \( y = 0 \):
\[
0 = \sqrt{x + 4} \implies x + 4 = 0 \implies x = -4 \implies (-4, 0)
\]

3. **Vertical Asymptote**: Not applicable since it's a square root function.

4. **Range**: The minimum value occurs at \( x = -4 \), giving \( y \geq 0 \); thus the range is \( [0, \infty) \).

### Summary of Results:
- A: Linear function with intercepts (0, 12), (48, 0), range \( (-\infty, \infty) \)
- B: Quadratic with intercepts (0, 0), range \( [0, \infty) \)
- C: Quadratic with intercepts at \( (3+\sqrt{2}, 0) \) and \( (3-\sqrt{2}, 0) \), range \( [-2, \infty) \)
- D: Rational with vertical asymptote at \( x = 1 \), intercept (0, 0), range \( (-\infty, 2) \cup (2, \infty) \)
- E: Rational with vertical asymptotes at \( x = 3 \) and \( x = -3 \), intercept (0, -1/9), range \( (-\infty, 0) \)
- F: Square root with intercepts (0, 2) and (-4, 0), range \( [0, \infty) \)