Scale input and output distances

Scale coordinates, using the reciprocal of the replacement multiplier

Start by deciding which coordinate changes. A multiplier inside f changes horizontal distances; a multiplier outside f changes vertical distances. Use a point to check the factor instead of guessing from the equation.

For the replacement rules below, $a>0$ and $(p,q)$ lies on $y=f(x)$.

Scale horizontal distances

Replace x by $ax$ to get $y=f(ax)$. To reproduce the old output at input p, solve $ax=p$. The new input is $p/a$, so the point map is $(p,q)\mapsto(p/a,q)$.

The scale factor is $1/a$, parallel to the x-axis. Thus $f(2x)$ halves horizontal distances from the y-axis; it does not double them. The y-axis stays fixed.

Scale vertical distances

Replace y by $ay$ to get $ay=f(x)$, hence $y=f(x)/a$. The point map is $(p,q)\mapsto(p,q/a)$: scale by $1/a$ parallel to the y-axis, keeping the x-axis fixed.

Be careful about the equation you are given: $2y=f(x)$ halves y-coordinates, while $y=2f(x)$ doubles them.

Check asymptotes and intercepts

• Example 3 halves x-coordinates. Its vertical asymptote moves from $x=-1$ to $x=-1/2$; $y=1$ stays fixed.
• Example 4 doubles y-coordinates. Its horizontal asymptote moves from $y=1$ to $y=2$; $x=-1$ stays fixed.

Apply the same replacement to each asymptote equation before solving it. In Self-Practice 2, $2f(x)$ and $f(2x)$ happen to give lines with equal slopes, but their y-intercepts differ. Equal slopes do not make the graphs identical.

Check before the examples: if $(4,3)$ is on $y=f(x)$, the corresponding point is $(2,3)$ on $y=f(2x)$ and $(4,6)$ on $y=2f(x)$.

Scaling activity and source notes

The quadratic activities include every Teacher minimum/intercept table entry. Their phrase “shape preserved” means the graphs remain parabolas: scaling changes width or height and need not preserve distances or size. Unlike a translation, it fixes the relevant coordinate axis. Self-Practice 2’s answers are independently derived, rather than printed completions.