Given: $$f(x) = \frac{3}{x - 1} - 2$$ 3.1 Write down the equation of the: 3.1.1 horizontal asymptote of $f$ - NSC Mathematics - Question 3 - 2016 - Paper 1

The vertical asymptote occurs where the denominator is zero. Setting the denominator $x - 1 = 0$, we find:

$x - 1 = 0 \Rightarrow x = 1$

Thus, the equation of the vertical asymptote is:

$x = 1$

To find the $x$-intercept, we set $f(x) = 0$:

$\frac{3}{x - 1} - 2 = 0 \Rightarrow \frac{3}{x - 1} = 2 \Rightarrow 3 = 2(x - 1) \Rightarrow 3 = 2x - 2 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}$

Thus, the $x$-intercept is $

\left( \frac{5}{2}, 0 \right)$.

To find the $y$-intercept, we evaluate $f(0)$:

$f(0) = \frac{3}{0 - 1} - 2 = -3 - 2 = -5$

Thus, the $y$-intercept is $(0, -5)$.

To sketch the graph of $f(x)$, we will plot the horizontal asymptote $y = -2$ and the vertical asymptote $x = 1$. The $x$-intercept is at $(\frac{5}{2}, 0)$, and the $y$-intercept is at $(0, -5)$. The function approaches $y = -2$ as $x \to \infty$ and $x \to -\infty$, and the graph will exhibit a hyperbolic shape arising from the asymptotes.

$g(x) = f(x - 3) + 7$ implies that the vertical asymptote of $g$ is shifted to $x = 4$ (since $1 + 3 = 4$). The horizontal asymptote is lifted to $y = 5$ (since $-2 + 7 = 5$). Thus, the coordinates of the intersection of the asymptotes are:

$(4, 5)$.