2.9 Graphs, the modulus function, asymptotes and intersections — Practise

Work through these on your own. For a modulus equation, split into cases and reject any value that does not fit its case. Write an inequality solution as a set, and an asymptote as an equation.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Solve $|x - 3| = 2$ .
Solve $|2x - 1| = 5$ .
Solve $|3x - 1| = 4$ .
Solve $|x + 3| = |2x|$ .
Solve $|x| = 2x - 1$ .
Solve $|x + 1| < 3$ .
Solve $|x - 2| \geq 1$ .
Find the points where the curve $y = x^2$ meets the line $y = x + 6$ .
Find where the curve $y = x^2 - 4$ meets the line $y = 0$ , and explain what these points are.
Solve $\dfrac{1}{x} = 2$ .
Find the points where $y = \dfrac{1}{x}$ meets the line $y = x$ .
The curve $y = \dfrac{3}{x - 2} + 1$ has a horizontal asymptote. Find its equation by considering what happens as $x \to \infty$ .

Answers and concise solutions

1. The solutions

Answer: The expression inside the modulus is either $2$ or $-2$ . Case 1: $x - 3 = 2$ , so $x = 5$ . Case 2: $x - 3 = -2$ , so $x = 1$ . Both values satisfy the equation.

2. The solutions

Answer: Case 1: $2x - 1 = 5$ , so $2x = 6$ and $x = 3$ . Case 2: $2x - 1 = -5$ , so $2x = -4$ and $x = -2$ . Both values work.

3. The solutions

Answer: Case 1: $3x - 1 = 4$ , so $x = \frac{5}{3}$ . Case 2: $3x - 1 = -4$ , so $3x = -3$ and $x = -1$ . Both values work.

4. The solutions

Answer: Two expressions have the same modulus when they are equal or when one is the negative of the other. Case 1: $x + 3 = 2x$ , so $x = 3$ . Case 2: $x + 3 = -2x$ , so $3x = -3$ and $x = -1$ . Both values satisfy the equation.

5. The solution

Answer: Case 1, $x \geq 0$ : $x = 2x - 1$ , so $x = 1$ . Case 2, $x < 0$ : $-x = 2x - 1$ , so $3x = 1$ and $x = \frac{1}{3}$ . The value $x = \frac{1}{3}$ is not in the case $x < 0$ , so it is rejected. The only solution is $x = 1$ .

6. The solution set

Answer: $|x + 1| < 3$ means the distance from $x$ to $-1$ is less than $3$ . So $-3 < x + 1 < 3$ . Subtract $1$ from every part: $-4 < x < 2$ , or $x \in (-4, 2)$ .

7. The solution set

Answer: The distance from $x$ to $2$ is at least $1$ , so $x$ lies at or beyond the two points $1$ and $3$ . This gives $x \leq 1$ or $x \geq 3$ , that is $x \in (-\infty, 1] \cup [3, \infty)$ .

8. The intersection points

Answer: Set $x^2 = x + 6$ : $x^2 - x - 6 = 0$ , so $(x - 3)(x + 2) = 0$ . The values are $x = 3$ and $x = -2$ . Substituting into $y = x^2$ gives the points $(3, 9)$ and $(-2, 4)$ .

9. The intersection points

Answer: The line $y = 0$ is the $x$ -axis. Setting $x^2 - 4 = 0$ gives $x^2 = 4$ , so $x = 2$ or $x = -2$ . The points are $(2, 0)$ and $(-2, 0)$ ; they are the roots of $x^2 - 4 = 0$ , read from the graph as the places where the curve crosses the $x$ -axis.

10. The solution

Answer: The graph of $y = \frac{1}{x}$ meets the line $y = 2$ once. Solve: multiply both sides by $x$ (which is not zero), giving $1 = 2x$ , so $x = \frac{1}{2}$ . The value $x = 0$ is excluded from the domain.

11. The intersection points

Answer: Set $\frac{1}{x} = x$ . Multiply both sides by $x$ : $1 = x^2$ , so $x = 1$ or $x = -1$ . The points are $(1, 1)$ and $(-1, -1)$ .

12. The asymptote

Answer: As $x \to \infty$ , the fraction $\frac{3}{x - 2}$ gets closer and closer to $0$ . So $y \to 1$ . The horizontal asymptote is $y = 1$ . There is also a vertical asymptote at $x = 2$ , where the denominator is zero and the curve is undefined.