Question: EXERCISE 3
(a) Given: $y = \frac{1}{x}$ and $y = \frac{5}{x}$
(1) Which graph has branches that have the furthest stretch away from the x-axis?
Explain.
(2) Sketch the graphs on the same set of axes.
(3) Now sketch the graph of $y = -\frac{5}{x}$ on the same set of axes.
(b) Given: $y = \frac{2}{x} + 2$
(1) Write down the equations of the vertical and horizontal asymptotes.
(2) Determine the coordinates of the x-intercept.
(3) Sketch the graph on a set of axes.
(c) Given: $y = -\frac{4}{x} - 1$
(1) Write down the equations of the vertical and horizontal asymptotes.
(2) Determine the coordinates of the x-intercept.
(3) Sketch the graph on a set of axes.
(d) Sketch the graph of $y = \frac{3}{x} + 2$ on a set of axes. Indicate the coordinates of the
x-intercept as well as the asymptotes.
(e) Match the equations on the left to the graphs on the right.
(1) $y = \frac{2}{x}$
(2) $y = \frac{4}{x}$
(3) $y = -\frac{3}{x}$
(4) $y = -\frac{3}{x} + 1$
1. **Problem Statement:**
We analyze and sketch various rational functions of the form $y = \frac{k}{x} + c$ and identify asymptotes, intercepts, and graph behavior.
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### (a) Given: $y = \frac{1}{x}$ and $y = \frac{5}{x}$
1. **Which graph has branches that stretch furthest from the x-axis?**
The graph with $y = \frac{5}{x}$ has branches that stretch further away from the x-axis because the numerator $5$ is larger than $1$, making the values of $y$ larger in magnitude for the same $x$.
2. **Sketch graphs of $y = \frac{1}{x}$ and $y = \frac{5}{x}$ on the same axes:**
Both are hyperbolas with vertical asymptote $x=0$ and horizontal asymptote $y=0$. The $y=\frac{5}{x}$ graph is steeper and farther from the x-axis.
3. **Sketch $y = -\frac{5}{x}$ on the same axes:**
This graph is the reflection of $y=\frac{5}{x}$ about the x-axis, so branches lie in quadrants II and IV.
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### (b) Given: $y = \frac{2}{x} + 2$
1. **Asymptotes:**
- Vertical asymptote: $x=0$ (denominator zero)
- Horizontal asymptote: $y=2$ (constant added)
2. **Find x-intercept:**
Set $y=0$:
$$0 = \frac{2}{x} + 2$$
$$\frac{2}{x} = -2$$
$$2 = -2x$$
$$x = \frac{2}{-2} = -1$$
So, x-intercept is $(-1, 0)$.
3. **Sketch graph:**
Hyperbola shifted up by 2 units, vertical asymptote at $x=0$, horizontal asymptote at $y=2$, crossing x-axis at $(-1,0)$.
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### (c) Given: $y = -\frac{4}{x} - 1$
1. **Asymptotes:**
- Vertical asymptote: $x=0$
- Horizontal asymptote: $y=-1$
2. **Find x-intercept:**
Set $y=0$:
$$0 = -\frac{4}{x} - 1$$
$$\frac{4}{x} = -1$$
$$4 = -x$$
$$x = -4$$
So, x-intercept is $(-4, 0)$.
3. **Sketch graph:**
Hyperbola reflected (negative numerator) and shifted down by 1 unit.
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### (d) Given: $y = \frac{3}{x} + 2$
1. **Asymptotes:**
- Vertical: $x=0$
- Horizontal: $y=2$
2. **Find x-intercept:**
Set $y=0$:
$$0 = \frac{3}{x} + 2$$
$$\frac{3}{x} = -2$$
$$3 = -2x$$
$$x = -\frac{3}{2}$$
So, x-intercept is $\left(-\frac{3}{2}, 0\right)$.
3. **Sketch graph:**
Hyperbola shifted up by 2 units.
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### (e) Match equations to graphs:
- (1) $y=\frac{2}{x}$: standard hyperbola with branches in quadrants I and III.
- (2) $y=\frac{4}{x}$: similar shape but branches further from x-axis than (1).
- (3) $y=-\frac{3}{x}$: reflection of (1) about x-axis, branches in quadrants II and IV.
- (4) $y=-\frac{3}{x} + 1$: same as (3) but shifted up by 1 unit.
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**Final answers:**
- (a1) $y=\frac{5}{x}$ stretches further from x-axis.
- (b1) Vertical asymptote: $x=0$, Horizontal asymptote: $y=2$.
- (b2) x-intercept: $(-1,0)$.
- (c1) Vertical asymptote: $x=0$, Horizontal asymptote: $y=-1$.
- (c2) x-intercept: $(-4,0)$.
- (d) Vertical asymptote: $x=0$, Horizontal asymptote: $y=2$, x-intercept: $\left(-\frac{3}{2},0\right)$.
- (e) Matches as explained above.