Mathematics • Year 9 • Unit 2 • Lesson 10
Sketch and Identify — Mixed Challenge
Pull together every parabola skill from Unit 2: sketch in 5 steps, identify the equation from features, reverse-engineer from intercepts. Catch one tricky mistake, then invent a parabola family with prescribed conditions.
1. Mixed problems
Each problem mixes the sketch and identify skills. Show working. 3 marks each
1.1 Sketch $y = (x - 2)^2 - 9$. Label the vertex, $y$-intercept, and both $x$-intercepts with coordinates.
1.2 Sketch $y = -2(x + 1)^2 + 8$. Label vertex (max), $y$-intercept, and both $x$-intercepts.
1.3 A parabola has vertex $(3, 2)$ and passes through $(5, 10)$. (a) Write the equation in vertex form. Show every step. (b) Verify by computing $y$ at $x = 1$ (use the symmetry, then check by substitution).
1.4 A downward-opening parabola has $x$-intercepts at $(-2, 0)$ and $(6, 0)$, and a maximum value of $y = 8$. (a) Find the axis of symmetry using the midpoint of the $x$-intercepts. (b) State the vertex. (c) Substitute one $x$-intercept to find $a$, and write the equation in vertex form.
1.5 A parabola passes through $(1, 0)$, $(5, 0)$ and $(3, -8)$. (a) Use the symmetry of the $x$-intercepts to state the axis and so the value of $h$. (b) Why is $(3, -8)$ the vertex? (Hint: $x = 3$ is on the axis.) (c) Sub one $x$-intercept to find $a$. (d) Write the equation in vertex form.
1.6 For each statement, decide TRUE or FALSE. If false, write the corrected statement. (a) "If you know the vertex of a parabola, you know its equation." (b) "A parabola with vertex on the $x$-axis has exactly one $x$-intercept." (c) "If two parabolas have the same vertex and the same $a$, they are the same parabola." (d) "The $y$-intercept of $y = a(x - h)^2 + k$ is always $(0, k)$."
2. Find the mistake
A classmate is asked to sketch $y = (x + 2)^2 - 9$ in five steps. Exactly two of their steps are wrong. Spot them, explain why, and fix them. 3 marks
Student's working for $y = (x + 2)^2 - 9$:
Step 1: $a = 1 > 0$, opens UP, standard width. ✓
Step 2: Vertex $(2, -9)$ (flipping the $+2$ inside the bracket). ✗
Step 3: $y$-intercept: sub $x = 0$: $y = (2)^2 - 9 = -5$, so $(0, -5)$. ✓
Step 4: $x$-intercepts: $(x + 2)^2 = 9 \Rightarrow x + 2 = 3 \Rightarrow x = 1$ (just one). ✗
Step 5: Draw a smooth U through the labelled points. ✓
(a) Which two steps are wrong?
(b) For each wrong step, explain in one sentence why the student's reasoning is mistaken.
(c) Write the correct version of each wrong step.
Stuck on Step 2? "Flipping" only works correctly when applied to $h$ via the bracket: $(x + 2) = 0 \Rightarrow x = -2$, NOT $x = 2$.3. Open-ended challenge — design a parabola from scratch
Many valid answers. Be creative but precise. 4 marks
3.1 Design a parabola that satisfies ALL these conditions at once:
• Opens DOWNWARD.
• Has its MAXIMUM value at $y = 6$.
• Has $x$-intercepts $4$ units apart.
• Passes through the point $(0, -2)$.
Step-by-step:
(i) From "max value $= 6$", state $k$.
(ii) The $x$-intercepts are $4$ apart, so they sit $2$ units left and $2$ units right of the axis. Use $(0, -2)$ and the equation $y = a(x - h)^2 + 6$ to find $h$ AND $a$ together. (Hint: you'll need to combine two pieces of information — the location of the $x$-intercepts and the point $(0, -2)$.)
(iii) Write the final equation.
(iv) Verify by computing $y$ at the two $x$-intercepts.
Bonus: Sketch the parabola, labelling vertex, $y$-intercept, and both $x$-intercepts.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sketch $y = (x - 2)^2 - 9$
Direction: up ($a = 1$). Vertex $(2, -9)$ (min). $y$-int: $(0 - 2)^2 - 9 = -5$, so $(0, -5)$. $x$-ints: $(x - 2)^2 = 9 \Rightarrow x - 2 = \pm 3 \Rightarrow x = 5$ or $x = -1$, so $(5, 0)$ and $(-1, 0)$. Smooth U through four labelled points.
1.2 — Sketch $y = -2(x + 1)^2 + 8$
Direction: down ($a = -2$), narrow ($|a| = 2 > 1$). Vertex $(-1, 8)$ (max). $y$-int: $-2(1)^2 + 8 = 6$, so $(0, 6)$. $x$-ints: $0 = -2(x + 1)^2 + 8 \Rightarrow (x + 1)^2 = 4 \Rightarrow x + 1 = \pm 2 \Rightarrow x = 1$ or $x = -3$, so $(1, 0)$ and $(-3, 0)$. Smooth narrow inverted U through four labelled points.
1.3 — Vertex $(3, 2)$ through $(5, 10)$
(a) $y = a(x - 3)^2 + 2$. Sub $(5, 10)$: $10 = a(2)^2 + 2 = 4a + 2 \Rightarrow 4a = 8 \Rightarrow a = 2$. Equation: $y = 2(x - 3)^2 + 2$. (b) Symmetry: $x = 5$ is $2$ right of axis $x = 3$; mirror at $x = 1$ should also give $y = 10$. Check by substitution: $y = 2(1 - 3)^2 + 2 = 2(4) + 2 = 10$. ✓ Matches.
1.4 — Downward, $x$-ints $(-2, 0)$ and $(6, 0)$, max $y = 8$
(a) Midpoint of $-2$ and $6$: $(-2 + 6)/2 = 2$. Axis $x = 2$. (b) Vertex $(2, 8)$ (max value at axis). (c) Sub $(6, 0)$ into $y = a(x - 2)^2 + 8$: $0 = a(4)^2 + 8 = 16a + 8 \Rightarrow 16a = -8 \Rightarrow a = -\tfrac{1}{2}$. Equation: $y = -\tfrac{1}{2}(x - 2)^2 + 8$.
1.5 — Passes through $(1, 0)$, $(5, 0)$, $(3, -8)$
(a) Axis: midpoint of $1$ and $5$ is $3$. So $h = 3$. (b) $(3, -8)$ sits ON the axis of symmetry, so it must be the vertex (the only point on the axis that's on the curve). $k = -8$. (c) Sub $(1, 0)$ into $y = a(x - 3)^2 - 8$: $0 = a(-2)^2 - 8 = 4a - 8 \Rightarrow 4a = 8 \Rightarrow a = 2$. (d) Equation: $y = 2(x - 3)^2 - 8$.
1.6 — True / False
(a) FALSE. Vertex alone fixes $h$ and $k$, but $a$ is still unknown — many parabolas share a vertex. Need vertex AND one other point.
(b) TRUE. Vertex on the $x$-axis means $k = 0$, so the only intercept is the vertex itself (one repeated root).
(c) TRUE. Same vertex pins down $h$ and $k$; same $a$ pins down the shape. Identical parabolas.
(d) FALSE. $y$-intercept is $y = a(0 - h)^2 + k = ah^2 + k$. Only equals $k$ when $h = 0$ (vertex on the $y$-axis).
2 — Find the mistake
(a) The two wrong steps are Step 2 and Step 4.
(b) Step 2: "$(x + 2)$" is zero when $x = -2$, NOT $x = 2$. The vertex's $x$-coordinate is the OPPOSITE of the sign you see in the bracket (so $h = -2$ here). Step 4: The student forgot the $\pm$ when taking the square root of $9$, losing one of the two intercepts.
(c) Step 2 corrected: Vertex $(-2, -9)$. (Bracket zero at $x = -2$.) Step 4 corrected: $(x + 2)^2 = 9 \Rightarrow x + 2 = \pm 3 \Rightarrow x = 1$ or $x = -5$. Two $x$-intercepts: $(1, 0)$ and $(-5, 0)$.
Note: this also fixes Step 3's $y$-intercept arithmetic, which was actually correct — $y = (0 + 2)^2 - 9 = 4 - 9 = -5$. So the only two wrong steps are 2 and 4.
3 — Open-ended challenge (sample solution)
(i) Max value $6 \Rightarrow k = 6$.
(ii) The $x$-intercepts are $h - 2$ and $h + 2$ (since they're $4$ units apart and symmetric about $x = h$). Sub one $x$-intercept, say $(h + 2, 0)$, into $y = a(x - h)^2 + 6$: $0 = a(2)^2 + 6 = 4a + 6 \Rightarrow a = -\tfrac{3}{2}$. Now sub the second condition $(0, -2)$ into the same equation with $a = -\tfrac{3}{2}$: $-2 = -\tfrac{3}{2}(0 - h)^2 + 6 \Rightarrow -8 = -\tfrac{3}{2}h^2 \Rightarrow h^2 = \tfrac{16}{3} \Rightarrow h = \pm\tfrac{4}{\sqrt{3}} \approx \pm 2.31$. Take either sign; for definiteness pick $h = \tfrac{4}{\sqrt{3}} \approx 2.31$.
(iii) Equation: $y = -\tfrac{3}{2}(x - 2.31)^2 + 6$ (approximately).
(iv) Verify $x$-intercepts: at $x = 2.31 + 2 = 4.31$, $y = -\tfrac{3}{2}(2)^2 + 6 = -6 + 6 = 0$. ✓ At $x = 2.31 - 2 = 0.31$, $y = -\tfrac{3}{2}(-2)^2 + 6 = 0$. ✓ And at $x = 0$: $y = -\tfrac{3}{2}(2.31)^2 + 6 \approx -\tfrac{3}{2}(5.33) + 6 \approx -8 + 6 = -2$. ✓ All three conditions satisfied.
Marking: 1 mark for $k = 6$; 1 mark for finding $a$ from the intercept-spacing condition; 1 mark for finding $h$ from the $(0, -2)$ condition; 1 mark for verification. Award full marks for any valid algebraic approach that produces a parabola meeting all four conditions (the negative $h$ solution is equally valid).