Given: $g(x) = \frac{1}{x-1} + 2$ 4.1 Write down the equations of the asymptotes of $g$ - NSC Mathematics - Question 4 - 2024 - Paper 1

To graph $g(x)$, we start by identifying key points:

1. X-intercept: Set $g(x) = 0$: $0 = \frac{1}{x-1} + 2$ $\Rightarrow -2 = \frac{1}{x-1}$ $\Rightarrow 1 = -2(x-1) \Rightarrow x = -\frac{1}{2}$ Thus, the x-intercept is $(-\frac{1}{2}, 0)$.

2. Y-intercept: Set $x = 0$: $g(0) = \frac{1}{0-1} + 2 = -1 + 2 = 1$ Thus, the y-intercept is $(0, 1)$.

3. The graph approaches the asymptotes at $x = 1$ and $y = 2$. The shape of $g$ indicates that the function decreases from the top left to the right, crossing the y-axis and approaching the horizontal asymptote as $x$ increases.

To find where $g(x) > 0$, we solve the inequality:

1. Start from the function: $g(x) = \frac{1}{x-1} + 2 > 0$ Rearranging gives: $\frac{1}{x-1} > -2$ This can be expressed as: $1 > -2(x-1)$ $1 > -2x + 2$ $2x > 1 \Rightarrow x > \frac{1}{2}$

2. Since $g(x)$ is undefined at $x = 1$, we must also consider that interval: The solution is: $x \in (1, \infty)$, as $g(x)$ is positive for $x > 1$.

To determine the axis of symmetry, we first recognize that rational functions of this form often have symmetry around their vertical asymptote. Our vertical asymptote is at $x = 1$.

Thus, the general equation of symmetry can be expressed as: $y = -x + c$ Where $c$ can be found by substituting any point on the function. Using the point of intersection of the asymptotes as a reference:

For $y = 2$ when $x = 1$:

• We substitute: $2 = -1 + c \Rightarrow c = 3$

Therefore, the equation of the axis of symmetry with a negative gradient is: $y = -x + 3$.