How to sketch transformed graphs from asymptotes and intercepts

Key takeaway: When a graph is transformed, the key is to track what happens to asymptotes, intercepts, and the overall shape. Here the three expressions change the graph in three different ways: horizontal scaling and shifting, reflection and vertical stretch, and reciprocal transformation.

(i) $y=f\left(\frac12 x\right)-1$

This is a horizontal stretch by factor 2 and then a shift down 1.

• The vertical asymptote $x=-2$ becomes $x=-4$.
• The slant asymptote $y=2x-4$ changes as follows:
  • first replace $x$ by $\frac12 x$, so the asymptote becomes $y=2\left(\frac12 x\right)-4=x-4$
  • then shift down 1: $y=x-5$
• The $x$-intercepts move from $x=-3$ and $x=4$ to twice those values: $x=-6 \text{ and } x=8$ and then down 1, so the new curve passes through $(-6,-1)$ and $(8,-1)$.

(ii) $y=-2f(x+3)$

This is a shift left 3, then a reflection in the $x$-axis and a vertical stretch by factor 2.

• The vertical asymptote $x=-2$ becomes $x=-5$.
• The slant asymptote transforms by substituting $x+3$ into $f$: $y=-2\big(2(x+3)-4\big)=-2(2x+2)=-4x-4$
• The original zeros at $x=-3$ and $x=4$ shift to $x=-6$ and $x=1$. Because of the factor $-2$, the corresponding points lie on the $x$-axis still, since $f(x)=0$ gives $y=0$.

(iii) $y=\frac1{f(x)}$

For the reciprocal graph:

• Wherever $f(x)=0$, the new graph has vertical asymptotes. So vertical asymptotes occur at $x=-3 \text{ and } x=4$
• Wherever $f(x)$ has a vertical asymptote, the reciprocal tends to $0$. So $x=-2$ becomes a horizontal asymptote candidate.
• Since the original slant asymptote is $y=2x-4$, the reciprocal approaches $0$ as $|x|\to\infty$.

So the sketch should show:

• vertical asymptotes at $x=-3$ and $x=4$,
• the graph approaching $y=0$ for large $|x|$,
• and sign changes depending on whether $f(x)$ is positive or negative on each interval.

A good sketch is built interval by interval: $(-\infty,-3)$, $(-3,-2)$, $(-2,4)$, and $(4,\infty)$.

---

Pitfalls the pros know 👇 A common mistake is to treat the transformations in the wrong order. For $f\left(\frac12 x\right)$, the graph stretches horizontally by factor 2; it does not shrink. Another easy slip is to forget that zeros of $f$ become vertical asymptotes in $1/f(x)$, while asymptotes of $f$ usually turn into zeros or limiting lines for the reciprocal.

What if the problem changes? If the expression were $y=f(2x)-1$, the horizontal effect would reverse: the graph would compress toward the $y$-axis by factor $\tfrac12$. If it were $y=\frac{1}{f(x)+1}$, then the points where $f(x)=-1$ would become vertical asymptotes instead of the zeros of $f$.

Tags: vertical asymptote, horizontal stretch, reciprocal function