Mathematics • Year 9 • Unit 2 • Lesson 16
Graphing Non-Linear — Mixed Challenge
Six mixed graphing problems pulling together all three families, one "find the mistake" with a classic plotting error, and an open-ended challenge: design your own graph with prescribed features.
1. Mixed graphing problems
Choose the right $x$-values, build the table, pick a scale, describe the curve. Show your working. 3 marks each
1.1 Build a table for $y = x^2 - 3$ at $x = -3, -2, -1, 0, 1, 2, 3$. State the $y$-intercept, the two $x$-intercepts (read from the table) and the vertex.
1.2 Build a table for $y = \dfrac{-6}{x}$ at $x = \pm 1, \pm 2, \pm 3, \pm 6$. State which two quadrants the branches sit in (and WHY).
1.3 Build a table for $y = \left(\tfrac{1}{2}\right)^x$ at $x = -2, -1, 0, 1, 2, 3$. State the equation of the horizontal asymptote and whether the curve is growth or decay.
1.4 On the same set of axes you plan to plot $y = x^2$ and $y = 2^x$ for $x = 1, 2, 3, 4$. (a) Build a combined table. (b) At which integer $x$-value(s) do the two curves take the same $y$-value? (c) Beyond that, which curve grows faster?
1.5 A circle has equation $x^2 + y^2 = 25$. (a) State the centre, the radius, and which axis intercepts the circle passes through. (b) Explain in one sentence why a table-of-values approach is awkward for circles (compared to parabolas).
1.6 Sort the equations into families (parabola / hyperbola / exponential / circle / linear), then state ONE key plotting decision for each: $y = x^2 + 4$, $y = \tfrac{10}{x}$, $y = 2^x$, $x^2 + y^2 = 9$, $y = 3x - 1$, $y = (x - 2)^2$, $y = \tfrac{-3}{x}$.
2. Find the mistake
A student attempted to graph $y = \dfrac{12}{x}$ using the table below and made three errors of judgment. Identify each mistake, explain why it's wrong, and write the corrected step. 3 marks
Student's working:
$x$-values chosen: $0, 1, 2, 3$ (four values only, all positive).
At $x = 0$: $y = \dfrac{12}{0} = 0$ (entered as a point).
At $x = 1, 2, 3$: $y = 12, 6, 4$.
Plotted $(0,0), (1,12), (2,6), (3,4)$, joined with straight line segments.
(a) Identify the three mistakes.
(b) Explain each in one sentence.
(c) Describe the correct approach for graphing $y = \dfrac{12}{x}$ from $-12$ to $12$.
Stuck? Three errors to find: domain, point count, join method. Lesson § "Spot the Trap" and § "Common Pitfalls" both cover this.3. Open-ended challenge — design a graph
This question has many valid answers. Be precise. 4 marks
3.1 Design THREE non-linear equations that ALL pass through the point $(2, 4)$. The three must come from THREE different families (one parabola, one hyperbola, one exponential).
For each:
(i) Write the equation.
(ii) Verify $(2, 4)$ satisfies it.
(iii) Build a small table of $4$–$5$ values around $x = 2$ (e.g. $x = 0, 1, 2, 3, 4$).
(iv) State ONE key feature you'd label on the graph (vertex, asymptote, $y$-intercept).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $y = x^2 - 3$
Table: $x = -3, -2, -1, 0, 1, 2, 3$ gives $y = 6, 1, -2, -3, -2, 1, 6$. $y$-intercept: $(0, -3)$. Vertex: $(0, -3)$. $x$-intercepts are between $-2$ and $-1$, and between $1$ and $2$ (not at integers in this table — exact values $x = \pm \sqrt{3} \approx \pm 1.73$).
1.2 — $y = \dfrac{-6}{x}$
Table: $x = -6, -3, -2, -1, 1, 2, 3, 6$ gives $y = 1, 2, 3, 6, -6, -3, -2, -1$. Branches sit in Q2 (negative $x$, positive $y$) and Q4 (positive $x$, negative $y$) because $k = -6 < 0$ flips the standard $k > 0$ pattern across the axes.
1.3 — $y = (\tfrac{1}{2})^x$
Table: $x = -2, -1, 0, 1, 2, 3$ gives $y = 4, 2, 1, \tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}$. Asymptote: $y = 0$. This is exponential DECAY because base $\tfrac{1}{2}$ is between $0$ and $1$ — the curve decreases (left-to-right) towards the asymptote.
1.4 — $y = x^2$ vs $y = 2^x$
(a) $x = 1, 2, 3, 4$ gives $y = x^2$: $1, 4, 9, 16$ and $y = 2^x$: $2, 4, 8, 16$.
(b) Same $y$-value at $x = 2$ (both $= 4$) and $x = 4$ (both $= 16$).
(c) Between $x = 2$ and $x = 4$, the parabola is higher; beyond $x = 4$ the exponential takes over for good. Exponential growth EVENTUALLY beats polynomial growth.
1.5 — Circle $x^2 + y^2 = 25$
(a) Centre $(0, 0)$, radius $\sqrt{25} = 5$. The circle passes through $(\pm 5, 0)$ on the $x$-axis and $(0, \pm 5)$ on the $y$-axis.
(b) A table-of-values approach is awkward for a circle because EACH $x$-value gives TWO $y$-values (upper and lower half: $y = \pm \sqrt{25 - x^2}$) — it's not a function. Better to use a compass at the centre with the radius set to $5$.
1.6 — Sort and key decisions
Parabola: $y = x^2 + 4$ (use symmetric integers, e.g. $x = -3$ to $3$), $y = (x - 2)^2$ (symmetric around the axis $x = 2$, e.g. $x = -1$ to $5$).
Hyperbola: $y = \tfrac{10}{x}$ (use factors of $10$, exclude $x = 0$), $y = \tfrac{-3}{x}$ (use factors of $3$, exclude $x = 0$; branches in Q2/Q4).
Exponential: $y = 2^x$ (use $x = -2$ to $3$; large $y$-scale needed).
Circle: $x^2 + y^2 = 9$ (radius $3$; compass not table).
Linear: $y = 3x - 1$ (only two points needed; ruler join, no curve).
2 — Find the mistake (three errors)
(a) The three mistakes are: (i) including $x = 0$ in the table; (ii) only using four positive $x$-values (the negative branch is missing); (iii) joining the points with straight line segments instead of a smooth curve.
(b) Explanations: (i) $\tfrac{12}{0}$ is UNDEFINED — there is no point at $x = 0$, so it shouldn't be plotted (and certainly not as $y = 0$); (ii) hyperbolas have TWO branches, one for $x > 0$ and one for $x < 0$ — leaving out negative $x$-values misses half the curve; (iii) non-linear curves must be joined with smooth curves, never straight segments — "non-linear" literally means "not a line".
(c) Correct approach: use $x = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ (factors of $12$, both signs, exclude $0$); compute $y$ for all $12$ values; plot all points; join the six positive-$x$ points with one smooth branch in Q1 and the six negative-$x$ points with another smooth branch in Q3; label the asymptotes $x = 0$ and $y = 0$.
3 — Open-ended challenge (sample solutions)
Parabola example: $y = x^2$. Check: $(2)^2 = 4$ ✓. Table $x = 0, 1, 2, 3, 4$ gives $y = 0, 1, 4, 9, 16$. Key feature: vertex $(0, 0)$.
Hyperbola example: $y = \tfrac{8}{x}$. Check: $\tfrac{8}{2} = 4$ ✓. Table $x = 1, 2, 4, 8$ (skip $0$) gives $y = 8, 4, 2, 1$. Key feature: asymptotes $x = 0$ and $y = 0$.
Exponential example: $y = 2^x$. Check: $2^2 = 4$ ✓. Table $x = 0, 1, 2, 3, 4$ gives $y = 1, 2, 4, 8, 16$. Key feature: $y$-intercept $(0, 1)$; asymptote $y = 0$.
Marking: 1 mark per valid equation+verification+table+labelled feature, plus 1 mark for using THREE different families. Award full marks for any three correct, distinct equations.