A curve $y=f(x)$ undergoes in succession, the following transformations: A: A reflection in the y-axis B: A translation of 2 units in the direction of the x-axis C: A translation of $-a$ units in the direction of the y-axis, where $a>1$

1. Determine the expression for the curve after applying transformations A, B, and C sequentially to $f(x)$.

Let the initial curve be $y = f(x)$.
Transformation A (reflection in the $y$-axis): Replace $x$ with $-x$ to get $y = f(-x)$.
Transformation B (translation of $2$ units in the positive $x$-direction): Replace $x$ with $(x-2)$ to get $y = f(-(x-2)) = f(2-x)$.
Transformation C (translation of $-a$ units in the $y$-direction): Subtract $a$ from the function to get $y = f(2-x) - a$.

2. Equate the resulting expression to $g(x)$ and isolate the function term.

Since the resulting curve is $y = g(x)$, we have:
$f(2-x) - a = \frac{1-3a}{-x+5}$
$f(2-x) = \frac{1-3a}{-x+5} + a$

3. Use substitution to find the original function $f(x)$ and simplify.

Let $u = 2-x$, which implies $x = 2-u$.
Substitute $x = 2-u$ into the equation:
$f(u) = \frac{1-3a}{-(2-u)+5} + a$
$f(u) = \frac{1-3a}{u+3} + a$
Combine into a single fraction over a common denominator:
$f(u) = \frac{1-3a + a(u+3)}{u+3}$
$f(u) = \frac{1-3a + au + 3a}{u+3}$
$f(u) = \frac{au+1}{u+3}$
Therefore, replacing $u$ with $x$ yields $f(x) = \frac{ax+1}{x+3}$.

1. Determine the asymptotes, intercepts, and general shape of $y = g(x)$.

Given $g(x) = \frac{1-3a}{-x+5} = \frac{3a-1}{x-5}$ and $a > 1$.
Since $a > 1$, the constant numerator $3a-1 > 0$.
The vertical asymptote occurs where the denominator is zero: $x = 5$.
The horizontal asymptote is $y = 0$ (as $x \to \pm \infty$, $g(x) \to 0$).
To find the $y$-intercept, substitute $x = 0$: $y = \frac{1-3a}{5}$.
Since $a > 1$, $\frac{1-3a}{5} < 0$, so the $y$-intercept is negative.
There is no $x$-intercept since the numerator $1-3a \neq 0$.
The curve is a rectangular hyperbola with two branches: one in the region $x > 5, y > 0$ and the other in $x < 5, y < 0$.

2. Determine the asymptotes and intercepts of $y = g^{-1}(x)$ using its reflection property.

The graph of $y = g^{-1}(x)$ is the reflection of the graph of $y = g(x)$ in the line $y = x$.
Reflecting the asymptotes of $g(x)$ gives the asymptotes of $g^{-1}(x)$:

• The vertical asymptote is $x = 0$ (the $y$-axis).
• The horizontal asymptote is $y = 5$.
  Reflecting the $y$-intercept of $g(x)$ gives the $x$-intercept of $g^{-1}(x)$, which is at $x = \frac{1-3a}{5}$.
  Similarly, since $g(x)$ has no $x$-intercept, $g^{-1}(x)$ has no $y$-intercept.
  The curve is a rectangular hyperbola with two branches: one in the region $x > 0, y > 5$ and the other in $x < 0, y < 5$.

3. State the sketching characteristics for both graphs.

Sketch construction steps:

1. Draw the $x$ and $y$ axes and the dashed lines for the asymptotes $x = 5$ and $y = 5$.
2. For $y = g(x)$, draw the two branches approaching $x=5$ and $y=0$, ensuring it crosses the $y$-axis at $(0, \frac{1-3a}{5})$.
3. For $y = g^{-1}(x)$, draw the two branches approaching $x=0$ and $y=5$, ensuring it crosses the $x$-axis at $(\frac{1-3a}{5}, 0)$.
4. Ensure the graphs are symmetrical about the line $y = x$, and that they intersect each other exactly on this line (one intersection in the positive quadrant, one in the negative quadrant).