Mathematics • Year 9 • Unit 2 • Lesson 9
Intercepts of Parabolas
Build the intercept routine: sub $x = 0$ for the $y$-intercept, set $y = 0$ for the $x$-intercepts, remember the $\pm$ when square-rooting. 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. Find the $y$-intercept and the $x$-intercepts of $y = (x - 1)^2 - 9$.
Step 1 — $y$-intercept: substitute $x = 0$.
$y = (0 - 1)^2 - 9 = 1 - 9 = -8$. Point $(0, -8)$.
Reason: the $y$-intercept is where the curve crosses the $y$-axis, which always happens at $x = 0$.
Step 2 — $x$-intercepts: set $y = 0$.
$0 = (x - 1)^2 - 9 \Rightarrow (x - 1)^2 = 9$.
Reason: the $x$-intercepts are where the curve crosses the $x$-axis, so $y$ must equal $0$ there.
Step 3 — Take the square root (with $\pm$!).
$x - 1 = \pm 3$. So $x = 1 + 3 = 4$ or $x = 1 - 3 = -2$.
Reason: BOTH $(+3)^2 = 9$ AND $(-3)^2 = 9$, so the equation has two solutions. Forgetting the $\pm$ loses half the answer.
Step 4 — Write the points.
$x$-intercepts: $(4, 0)$ and $(-2, 0)$.
Sanity check: vertex is $(1, -9)$ (below the $x$-axis); curve opens up; so two crossings as expected.
Answer: $y$-intercept $(0, -8)$; $x$-intercepts $(4, 0)$ and $(-2, 0)$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill each blank. 4 marks
Problem. Find the $y$-intercept and the $x$-intercepts of $y = x^2 - 4$.
Step 1 — $y$-intercept: Sub $x = 0$: $y = (0)^2 - 4 = $ __________ . Point $(0, $__________$)$.
Step 2 — Set $y = 0$: $0 = x^2 - 4 \Rightarrow x^2 = $ __________ .
Step 3 — Take $\pm$ square root: $x = \pm$ __________ .
Step 4 — Write the points: $x$-intercepts $($__________$, 0)$ and $($__________$, 0)$.
Sanity check: Vertex is at $($__________ , __________$)$ — below the $x$-axis. Curve opens __________ (UP / DOWN). So __________ $x$-intercepts expected. ✓
3. You do — independent practice
Foundation: one intercept type. Standard: both intercepts together. Extension: predict count without solving, or solve a tricky vertex-form.
Foundation — single intercept
3.1 Find the $y$-intercept of $y = (x - 2)^2 + 5$. 1 mark
3.2 Find the $x$-intercepts of $y = x^2 - 16$. Show working. 1 mark
3.3 How many $x$-intercepts does $y = 3x^2 + 1$ have? Justify in one sentence. 1 mark
3.4 Solve $(x - 4)^2 = 25$ for the $x$-intercepts of $y = (x - 4)^2 - 25$. 1 mark
Standard — both intercepts together
3.5 For $y = x^2 - 25$, find: (a) the $y$-intercept; (b) the $x$-intercepts (show working). 2 marks
3.6 Find the $x$-intercepts and $y$-intercept of $y = (x + 2)^2 - 16$. Show all working. 2 marks
Extension — predict and solve mixed
3.7 Without solving algebraically, decide how many $x$-intercepts each parabola has. Briefly justify each using direction and vertex. (a) $y = -2(x - 1)^2 + 8$ (b) $y = (x + 5)^2$ (c) $y = 4x^2 + 7$. 3 marks
3.8 Find the $x$-intercepts of $y = 2(x - 3)^2 - 18$. (Hint: isolate the square first, then take $\pm$ root, then add 3.) 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = x^2 - 4$)
Step 1: $y = (0)^2 - 4 = \mathbf{-4}$. Point $(0, \mathbf{-4})$.
Step 2: $x^2 = \mathbf{4}$.
Step 3: $x = \pm \mathbf{2}$.
Step 4: $x$-intercepts $(\mathbf{2}, 0)$ and $(\mathbf{-2}, 0)$.
Sanity check: Vertex $(\mathbf{0}, \mathbf{-4})$, opens UP, expect 2 $x$-intercepts. ✓
3.1 — $y$-intercept of $y = (x - 2)^2 + 5$
Sub $x = 0$: $y = (-2)^2 + 5 = 4 + 5 = 9$. $(0, 9)$.
3.2 — $x$-intercepts of $y = x^2 - 16$
$0 = x^2 - 16 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$. Points $(4, 0)$ and $(-4, 0)$.
3.3 — How many $x$-ints for $y = 3x^2 + 1$?
$a = 3 > 0$ (vertex is a min) and $c = 1 > 0$ (vertex above $x$-axis). Same signs $\Rightarrow$ 0 $x$-intercepts. The curve stays above the $x$-axis everywhere.
3.4 — $(x - 4)^2 = 25$
$x - 4 = \pm 5 \Rightarrow x = 4 + 5 = 9$ or $x = 4 - 5 = -1$. Points $(9, 0)$ and $(-1, 0)$.
3.5 — $y = x^2 - 25$
(a) $y$-intercept: sub $x = 0$, $y = -25$. $(0, -25)$. (b) $x$-intercepts: $x^2 = 25 \Rightarrow x = \pm 5$. Points $(5, 0)$ and $(-5, 0)$.
3.6 — $y = (x + 2)^2 - 16$
$y$-intercept: sub $x = 0$, $y = (2)^2 - 16 = 4 - 16 = -12$. $(0, -12)$. $x$-intercepts: $0 = (x + 2)^2 - 16 \Rightarrow (x + 2)^2 = 16 \Rightarrow x + 2 = \pm 4 \Rightarrow x = 2$ or $x = -6$. Points $(2, 0)$ and $(-6, 0)$.
3.7 — Predict the count
(a) Vertex $(1, 8)$ ABOVE $x$-axis; opens DOWN ($a = -2$). Opposite of "same side"; the inverted U must come down through the $x$-axis: 2 $x$-intercepts.
(b) Vertex $(-5, 0)$ sits ON the $x$-axis; opens up. Exactly 1 $x$-intercept (repeated, at $x = -5$).
(c) Vertex $(0, 7)$ ABOVE $x$-axis; opens UP ($a = 4$). Same side as the opening: 0 $x$-intercepts.
3.8 — $y = 2(x - 3)^2 - 18$
$0 = 2(x - 3)^2 - 18 \Rightarrow 2(x - 3)^2 = 18 \Rightarrow (x - 3)^2 = 9 \Rightarrow x - 3 = \pm 3 \Rightarrow x = 6$ or $x = 0$. Points $(6, 0)$ and $(0, 0)$.