Discussion questions · Section 1 of 2

Explain how the graph of $y=2/(3x-1)$ may be obtained from $y=1/x$ by appropriate transformations.

The supplied graph is $y=f(x)+3$, with maximum (−1,5), y-intercept 3 and a right-hand horizontal asymptote y=1.

Sketch on separate diagrams:

(i) $y=f(x)$;

(ii) $y=-f(x)+3$;

(iii) $y=f(-x+3)$;

(iv) $y=f(2x+1)$.

Clearly indicate asymptotes, turning points and intercepts where possible.

The given graph $y=f(x)$ shows A=(k,1), B=(2k,−2), C=(3k,1), with k>0 as noted in the solution.

Sketch $y=f(x/2+k)-1$, indicating the coordinates corresponding to A, B, C.

The original $y=f(x)$ undergoes, in order:

A:translation 4 units in the positive x-direction;

B:reflection in the y-axis;

C:stretch parallel to x by 2.

The resulting equation is $y=(x-1)/(x+1)$. Obtain the original equation.

Given $F(x)=(2x-x^2)/(x^2-2x-3)$, find:

(i) the asymptotes;

(ii) where the curve cuts the axes;

(iii) the stationary point.

Hence sketch F and use it to obtain separate graphs of $y=|F(x)|$ and $y=(2|x|-x^2)/(x^2-2|x|-3)=F(|x|)$.

Given the three supplied graphs of $y=f(x)$, sketch $y=1/f(x)$ on separate diagrams.

(a) Zeros x=1,2, y-intercept 2, minimum (3/2,−1/4).

(b) Vertical asymptote x=−1, horizontal asymptote y=−1, y-intercept 1, zero x=1.

(c) Vertical asymptote x=−1, zeros x=±5/2, y-intercept 2, and the oblique guide shown through (0,1), (2,0).