Question

1. consider the function $y = \frac{4}{x - 2}$. 8c rearrange $y = \frac{4}{x - 2}$ to make $x$ the subject. 8d hence state the range of the function. $(-\infty, 0) \cup (0, \infty)$ $(-\infty, 4) \cup (4, \infty)$ $(-\infty, 2) \cup (2, \infty)$

Explanation:

Step1: Identify function restriction

The function is $y=\frac{4}{x-2}$. The denominator $x-2
eq0$, so $x
eq2$.

Step2: Analyze possible y-values

As $x$ approaches 2 from left/right, $y\to-\infty$ or $y\to+\infty$. For any $y
eq0$, solve $x=\frac{4}{y}+2$, which is defined (since $y
eq0$ gives valid $x
eq2$). $y=0$ would require $4=0$, impossible.

Step3: State the range

All real numbers except 0, so $(-\infty, 0)\cup(0, \infty)$.

Answer:

A. $(-\infty, 0)\cup(0, \infty)$