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

1. $|-4| = 4$ and $|7| = 7$.
2. $x^2 + x - 6 = (x + 3)(x - 2)$.
3. From the factorisation, $x = -3$ or $x = 2$.

If question 1 was hard, revisit the meaning of the modulus: it is a distance from zero, so it is never negative.

The expression inside the modulus is either $3$ or $-3$. Write two equations.

Case 1: $x + 5 = 3$. Case 2: $x + 5 = -3$.

Case 1:

Case 2:

Both values are solutions, so $x = -2$ or $x = -8$.

Write the two cases, then solve each linear equation.

$4x - 3 = 9$ gives one value, and $4x - 3 = -9$ gives the other.

Case 1:

Case 2:

Both values satisfy the equation.

The inequality says the distance from $x$ to $2$ is at most $4$. Write it as a double inequality.

$-4 \leq x - 2 \leq 4$. Add $2$ to every part.

Add $2$ to each part:

The solution set is $x \in [-2, 6]$.

A greater-than modulus splits into two separate inequalities, joined by "or".

$2x + 1 > 5$ or $2x + 1 < -5$. Solve each.

Split into two cases:

First case: $2x > 4$, so $x > 2$. Second case: $2x < -6$, so $x < -3$. Therefore

that is $x \in (-\infty, -3) \cup (2, \infty)$.

Set the two expressions for $y$ equal: $x^2 - 1 = 3$.

Solve $x^2 = 4$ for both values of $x$, then find each $y$.

Set the expressions equal:

The line is $y = 3$, so both points have $y = 3$. The graphs meet at $(2, 3)$ and $(-2, 3)$.

The vertical asymptote is where the denominator is zero. For the horizontal asymptote, ask what happens to $\frac{2}{x}$ as $x$ grows large.

For the meeting point, solve $\frac{2}{x} = 1$ by multiplying both sides by $x$.

The vertical asymptote is $x = 0$, where the curve is undefined. As $x \to \infty$, the fraction $\frac{2}{x} \to 0$, so the horizontal asymptote is $y = 0$.

For the meeting point:

The curve meets the line $y = 1$ at $(2, 1)$.