Mathematics • Year 9 • Unit 2 • Lesson 10
Sketch and Identify Parabolas
Build the 5-step sketch (direction → vertex → $y$-intercept → $x$-intercepts → smooth curve) AND the reverse skill (given vertex + one point, solve for $a$). One worked example, one guided fill-in, then eight independent problems.
1. I do — fully worked example
Read every line. Each step explains why we make the call, not just what the answer is.
Problem. Sketch $y = -(x - 2)^2 + 4$ using all 5 steps. Label vertex and both intercepts with coordinates.
Step 1 — Direction and width from $a$.
$a = -1$: opens DOWN, standard width.
Reason: negative $a$ flips the U; $|a| = 1$ matches $y = x^2$.
Step 2 — Plot the vertex.
$(h, k) = (2, 4)$. This is a MAX (because $a < 0$).
Reason: $(h, k)$ is read directly from vertex form.
Step 3 — $y$-intercept (sub $x = 0$).
$y = -(0 - 2)^2 + 4 = -4 + 4 = 0$. Point $(0, 0)$.
Reason: substitute $x = 0$ to find where the curve crosses the $y$-axis.
Step 4 — $x$-intercepts (set $y = 0$).
$-(x - 2)^2 + 4 = 0 \Rightarrow (x - 2)^2 = 4 \Rightarrow x - 2 = \pm 2 \Rightarrow x = 0$ or $x = 4$.
Reason: solve $y = 0$ to find the $x$-axis crossings. Don't forget the $\pm$!
Step 5 — Draw the smooth curve through all four labelled points.
Smooth inverted U through $(0, 0)$, vertex $(2, 4)$, $(4, 0)$.
Notice: the $y$-intercept and one $x$-intercept happen to coincide at the origin.
Answer: Inverted U; vertex $(2, 4)$ (max); $y$-int $(0, 0)$; $x$-ints $(0, 0)$ and $(4, 0)$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill each blank. 4 marks
Problem. Sketch $y = (x - 1)^2 - 4$ using all 5 steps. Label vertex and intercepts.
Step 1 — Direction and width: $a = $ __________ , opens __________________ (UP / DOWN), __________________ width.
Step 2 — Vertex: $(h, k) = ($__________ , __________$)$. This is a __________________ (MAX / MIN).
Step 3 — $y$-intercept: Sub $x = 0$: $y = (0 - 1)^2 - 4 = $__________ . Point $(0, $__________ $)$.
Step 4 — $x$-intercepts: $0 = (x - 1)^2 - 4 \Rightarrow (x - 1)^2 = $__________ $\Rightarrow x - 1 = \pm$ __________ $\Rightarrow x = $__________ or __________ . Points: $($__________$, 0)$ and $($__________$, 0)$.
Step 5 — Smooth curve: Draw a smooth __________________ (U / inverted U) through the four labelled points.
3. You do — independent practice
Foundation: read one feature off the equation. Standard: full sketch with all intercepts labelled. Extension: reverse-engineer (given vertex + one point, find $a$).
Foundation — read one feature
3.1 State the vertex and direction of $y = -(x - 3)^2 + 2$. 1 mark
3.2 Find the $y$-intercept of $y = 2(x - 1)^2 + 3$. Show working. 1 mark
3.3 A parabola has vertex $(0, 0)$ and passes through $(2, 12)$. Find $a$ (write the equation $y = a x^2$ and substitute). 1 mark
3.4 A parabola has vertex $(1, -4)$ and passes through $(3, 0)$. Find $a$ and write the equation. 1 mark
Standard — full sketches
3.5 Sketch $y = (x - 1)^2 - 4$ using all 5 steps. Label vertex, $y$-intercept, and both $x$-intercepts with coordinates. 3 marks
3.6 Sketch $y = -2(x + 1)^2 + 8$. Label vertex (max), $y$-intercept, and both $x$-intercepts. 3 marks
Extension — reverse-engineer the equation
3.7 A parabola has vertex $(3, 2)$ and passes through $(5, 10)$. (a) Write the equation with $a$ unknown using the vertex. (b) Substitute $(5, 10)$ to solve for $a$. (c) State the final equation. 2 marks
3.8 A parabola has vertex $(-1, 5)$, opens downward, and passes through $(1, -3)$. (a) Write the equation using the vertex (with unknown $a$). (b) Substitute $(1, -3)$ to find $a$. (c) State the equation and check the sign of $a$ matches "opens down". 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = (x - 1)^2 - 4$)
Step 1: $a = \mathbf{1}$, opens UP, standard width.
Step 2: Vertex $(\mathbf{1}, \mathbf{-4})$. This is a MIN.
Step 3: $y = (0 - 1)^2 - 4 = 1 - 4 = \mathbf{-3}$. Point $(0, \mathbf{-3})$.
Step 4: $(x - 1)^2 = \mathbf{4} \Rightarrow x - 1 = \pm \mathbf{2} \Rightarrow x = \mathbf{3}$ or $\mathbf{-1}$. Points $(\mathbf{3}, 0)$ and $(\mathbf{-1}, 0)$.
Step 5: Smooth U through the four labelled points.
3.1 — Vertex and direction of $y = -(x - 3)^2 + 2$
$a = -1$, $h = 3$, $k = 2$. Vertex $(3, 2)$, opens DOWN.
3.2 — $y$-intercept of $y = 2(x - 1)^2 + 3$
$y = 2(0 - 1)^2 + 3 = 2(1) + 3 = 5$. $(0, 5)$.
3.3 — Vertex origin, through $(2, 12)$
$y = a x^2$. Sub $(2, 12)$: $12 = a(4) \Rightarrow a = 3$. Equation: $y = 3x^2$.
3.4 — Vertex $(1, -4)$ through $(3, 0)$
$y = a(x - 1)^2 - 4$. Sub $(3, 0)$: $0 = a(2)^2 - 4 = 4a - 4 \Rightarrow a = 1$. Equation: $y = (x - 1)^2 - 4$.
3.5 — Sketch $y = (x - 1)^2 - 4$
Direction: up ($a = 1$). Vertex $(1, -4)$ (min). $y$-intercept: $y = (-1)^2 - 4 = -3$, so $(0, -3)$. $x$-intercepts: $(x - 1)^2 = 4 \Rightarrow x = 3$ or $-1$, so $(3, 0)$ and $(-1, 0)$. Smooth U through the four labelled points.
3.6 — Sketch $y = -2(x + 1)^2 + 8$
Direction: down ($a = -2$), narrow. Vertex $(-1, 8)$ (max). $y$-intercept: $y = -2(0 + 1)^2 + 8 = -2 + 8 = 6$, so $(0, 6)$. $x$-intercepts: $0 = -2(x + 1)^2 + 8 \Rightarrow (x + 1)^2 = 4 \Rightarrow x + 1 = \pm 2 \Rightarrow x = 1$ or $x = -3$, so $(1, 0)$ and $(-3, 0)$. Smooth narrow inverted U through the four labelled points.
3.7 — Vertex $(3, 2)$ through $(5, 10)$
(a) $y = a(x - 3)^2 + 2$. (b) Sub $(5, 10)$: $10 = a(2)^2 + 2 = 4a + 2 \Rightarrow 4a = 8 \Rightarrow a = 2$. (c) Equation: $y = 2(x - 3)^2 + 2$.
3.8 — Vertex $(-1, 5)$, opens down, through $(1, -3)$
(a) $y = a(x + 1)^2 + 5$. (b) Sub $(1, -3)$: $-3 = a(2)^2 + 5 = 4a + 5 \Rightarrow 4a = -8 \Rightarrow a = -2$. (c) Equation: $y = -2(x + 1)^2 + 5$. Check: $a = -2 < 0$ — yes, opens DOWN, matches the brief. ✓