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NSWMaths AdvancedSyllabus dot point

How do we sketch graphs built from sums, differences, products, quotients and reciprocals of standard functions?

Sketch graphs of sums, differences, products, quotients, squares and reciprocals of two known functions

A focused answer to the HSC Maths Advanced dot point on combining functions graphically. How to build sketches of $f + g$, $f g$, $f / g$, $1 / f$ and $f^2$ from the graphs of $f$ and $g$, where features come from, and what asymptotes and zeros do, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answer

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What this dot point is asking

NESA wants you to sketch a function built by combining two known functions. The combinations are sums f+gf + g, differences fβˆ’gf - g, products fgf g, quotients fg\frac{f}{g}, squares f2f^2, and reciprocals 1f\frac{1}{f}. You need to read features off the parent graphs and predict the combined graph.

The answer

Sums and differences

For y=f(x)+g(x)y = f(x) + g(x), add the heights of the two graphs at each xx. Useful checks:

  • Zeros of f+gf + g occur where the two graphs are reflections of each other through the xx-axis, that is f(x)=βˆ’g(x)f(x) = -g(x), not generally where either function is zero alone.
  • If ff is bounded and gg is unbounded, the long-term behaviour of f+gf + g is the same as gg.
  • For fβˆ’gf - g, subtract the heights. At points where f=gf = g, the difference is zero.

Products

For y=f(x)g(x)y = f(x) g(x), multiply the heights:

  • Zeros of fgf g are the union of the zeros of ff and gg.
  • The sign of fgf g follows the rule of signs: (+)(+)=+(+)(+) = +, (+)(βˆ’)=βˆ’(+)(-) = -, and so on.
  • If one factor is bounded between βˆ’1-1 and 11 (like sin⁑x\sin x), the other factor acts as an envelope: ∣fgβˆ£β‰€βˆ£f∣|f g| \le |f| and the graph oscillates between y=Β±f(x)y = \pm f(x).
  • If both factors grow, fgf g grows faster.

Quotients

For y=f(x)g(x)y = \frac{f(x)}{g(x)}:

  • Zeros of fg\frac{f}{g} are the zeros of ff, provided gg is non-zero there.
  • Vertical asymptotes occur at zeros of gg where ff is non-zero.
  • If g(x)=0g(x) = 0 and f(x)=0f(x) = 0 at the same point, there is a hole or a finite limit; check carefully.
  • Horizontal asymptotes come from the ratio of long-term behaviours: fgβ†’0\frac{f}{g} \to 0 if gg grows faster, a non-zero constant if they grow at the same rate, and ±∞\pm \infty if ff grows faster.

Reciprocals

The graph of y=1f(x)y = \frac{1}{f(x)} comes from y=f(x)y = f(x) by these rules:

  • Where f(x)=0f(x) = 0, 1f\frac{1}{f} has a vertical asymptote.
  • Where f(x)=Β±1f(x) = \pm 1, the reciprocal also equals Β±1\pm 1, so the two graphs meet on the lines y=1y = 1 and y=βˆ’1y = -1.
  • IMATH_50 has the same sign as ff everywhere fβ‰ 0f \neq 0.
  • Local maxima of ff where f>0f > 0 become local minima of 1f\frac{1}{f}, and vice versa, because reciprocating flips relative size.
  • As f(x)β†’Β±βˆžf(x) \to \pm \infty, 1f(x)β†’0\frac{1}{f(x)} \to 0. As f(x)β†’0Β±f(x) \to 0^{\pm}, 1f(x)β†’Β±βˆž\frac{1}{f(x)} \to \pm \infty.

Squares

For y=(f(x))2y = (f(x))^2:

  • Zeros of f2f^2 are the zeros of ff, but now they are double roots: the graph touches the xx-axis and turns.
  • IMATH_64 everywhere, so the graph never dips below the xx-axis.
  • Where f(x)=Β±1f(x) = \pm 1, f2(x)=1f^2(x) = 1. Where ∣f(x)∣>1|f(x)| > 1, f2>∣f∣f^2 > |f|. Where ∣f(x)∣<1|f(x)| < 1, f2<∣f∣f^2 < |f|.
  • Extrema of f2f^2 occur where ffβ€²=0f f' = 0, that is where f=0f = 0 or where ff has an extremum.

Worked examples

Reciprocal of IMATH_76

y=csc⁑x=1sin⁑xy = \csc x = \frac{1}{\sin x} has vertical asymptotes at x=kΟ€x = k \pi (zeros of sin⁑x\sin x), agrees with sin⁑x\sin x at sin⁑x=Β±1\sin x = \pm 1 (so at x=Ο€2+kΟ€x = \frac{\pi}{2} + k \pi), and is positive on intervals where sin⁑x>0\sin x > 0.

Square of IMATH_84

y=sin⁑2xy = \sin^2 x has zeros at x=kΟ€x = k \pi (double roots), touches the xx-axis there, is bounded above by 11, and equals 11 at x=Ο€2+kΟ€x = \frac{\pi}{2} + k \pi. Its period is Ο€\pi, half that of sin⁑x\sin x.

Quotient with shared zero

y=sin⁑xxy = \frac{\sin x}{x} has a hole at x=0x = 0 (because both numerator and denominator are zero there) with limit 11. Elsewhere it inherits zeros from sin⁑x\sin x at x=Β±Ο€,Β±2Ο€,…x = \pm \pi, \pm 2 \pi, \dots and decays in amplitude like 1∣x∣\frac{1}{|x|} as ∣xβˆ£β†’βˆž|x| \to \infty.

Product with growing envelope

y=eβˆ’xsin⁑xy = e^{-x} \sin x for xβ‰₯0x \ge 0 oscillates with the same zeros as sin⁑x\sin x (at x=kΟ€x = k \pi), but the amplitude decays. The envelopes are y=Β±eβˆ’xy = \pm e^{-x}, and the graph touches them where sin⁑x=Β±1\sin x = \pm 1.

Sum: line plus sine

y=x+sin⁑xy = x + \sin x has the line y=xy = x as a "spine" with small oscillations of amplitude 11 added. The graph is always within 11 of y=xy = x and crosses the line at every multiple of Ο€\pi.

Common traps

Treating reciprocals like reflections. 1f\frac{1}{f} is not a reflection. The shape distorts: large values become small and vice versa.

Forgetting asymptotes are about zeros of the denominator. For fg\frac{f}{g}, vertical asymptotes come from g(x)=0g(x) = 0, not f(x)=0f(x) = 0. Zeros come from f(x)=0f(x) = 0 (with g(x)β‰ 0g(x) \neq 0).

Square has zeros, not just minimums. f2f^2 has zeros wherever ff does. The graph touches the xx-axis at each one rather than crossing.

Misreading envelopes. For y=fgy = f g with ∣gβˆ£β‰€1|g| \le 1, the envelope is y=Β±fy = \pm f, not y=fy = f alone. Both branches matter.

Missing holes versus asymptotes in quotients. If ff and gg both vanish at the same point with a common factor, you get a hole, not an asymptote. Factor and cancel before concluding.

In one sentence

Combine known graphs feature by feature: sums add heights, products multiply heights (with the smaller factor acting as an envelope), quotients put zeros of the numerator on the xx-axis and zeros of the denominator at vertical asymptotes, reciprocals flip 0β†”βˆž0 \leftrightarrow \infty and meet at y=Β±1y = \pm 1, and squares fold everything above the xx-axis with double-root touches at the original zeros.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q143 marksThe function $f$ has zeros at $x = -2$ and $x = 3$ and is positive elsewhere. Describe the key features of the graph of $y = \frac{1}{f(x)}$.
Show worked answer β†’

Zeros of ff become vertical asymptotes of 1f\frac{1}{f}, so vertical asymptotes occur at x=βˆ’2x = -2 and x=3x = 3.

The sign of 1f\frac{1}{f} matches the sign of ff, so 1f\frac{1}{f} is positive on (βˆ’βˆž,βˆ’2)(-\infty, -2), (3,∞)(3, \infty), and wherever ff is positive in (βˆ’2,3)(-2, 3) (the whole interior if the question's "positive elsewhere" includes that interval, otherwise piecewise).

Local maxima of ff become local minima of 1f\frac{1}{f} in regions where f>0f > 0, and vice versa, because reciprocating flips relative size.

Markers reward identifying asymptotes from zeros, sign matching, and the inversion of extrema.

2021 HSC Q153 marksGiven $f(x) = x$ and $g(x) = \sin x$, sketch $y = f(x) g(x) = x \sin x$ for $-2\pi \le x \le 2\pi$.
Show worked answer β†’

Zeros of xsin⁑xx \sin x occur where x=0x = 0 or sin⁑x=0\sin x = 0, that is at x=0,Β±Ο€,Β±2Ο€x = 0, \pm \pi, \pm 2 \pi.

The amplitude grows linearly: ∣xsin⁑xβˆ£β‰€βˆ£x∣|x \sin x| \le |x|, with equality at sin⁑x=Β±1\sin x = \pm 1.

The envelope is y=Β±xy = \pm x, so the graph oscillates between y=βˆ’βˆ£x∣y = -|x| and y=∣x∣y = |x|, touching those lines at x=Ο€2+kΟ€x = \frac{\pi}{2} + k \pi.

Markers expect the zeros marked, the envelope drawn, and the oscillation shown with growing amplitude.

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