Mathematics • Year 9 • Unit 2 • Lesson 5
Vertical Translations — Mixed Challenge
Pull together every Lesson 5 idea — vertex from $c$, $y$-intercept = $c$, $x$-intercepts only when $c \le 0$ — with Lessons 2-4 reflexes (squaring, symmetry, opening direction). Catch a tricky intercept mistake, then take on an open-ended challenge.
1. Mixed problems
Show your working. Watch for negative roots and for cases with no $x$-intercepts. 3 marks each
1.1 For $y = x^2 - 9$: state the vertex, both intercepts (if they exist), and the range.
1.2 For each equation, state the vertex and whether real $x$-intercepts exist: (a) $y = x^2 + 1$ (b) $y = x^2$ (c) $y = x^2 - 1$ (d) $y = x^2 - 100$.
1.3 A parabola has the same shape as $y = x^2$, opens upward, and passes through $(2, 7)$. Find its equation, vertex, and $y$-intercept.
1.4 Build tables for $y = x^2$, $y = x^2 + 3$, and $y = x^2 - 2$ at $x = -2, -1, 0, 1, 2$. Use them to show that the three curves are vertical translations of each other.
1.5 Decide TRUE or FALSE with a one-line justification: (a) the vertex of $y = x^2 + c$ is $(0, c)$ for any $c$; (b) $y = x^2 - 4$ has two real $x$-intercepts; (c) $y = x^2 + 1$ has a $y$-intercept at $(0, 1)$; (d) the $y$-intercept of $y = x^2 + c$ is always positive.
1.6 The parabola $y = x^2 + c$ has $x$-intercepts at $(5, 0)$ and $(-5, 0)$. Find $c$ and state the vertex.
2. Find the mistake
Another student has tried to find the $x$-intercepts of $y = x^2 - 25$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why, then re-do the working correctly. 3 marks
Student's working — find the $x$-intercepts of $y = x^2 - 25$:
Line 1: Set $y = 0$: $0 = x^2 - 25$.
Line 2: Rearrange: $x^2 = 25$.
Line 3: Take the square root: $x = 5$.
Line 4: So the only $x$-intercept is $(5, 0)$.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including BOTH $x$-intercepts as ordered pairs.
Stuck? The parabola is symmetric across the $y$-axis — if $x = 5$ is on the curve at $y = 0$, what other value of $x$ also satisfies $x^2 = 25$?3. Open-ended challenge — design parabolas with target intercepts
This challenge has many valid answers. 4 marks
3.1 Using only the form $y = x^2 + c$ (no dilation, no reflection), design three different parabolas:
(i) One that has $x$-intercepts at $(\pm 6, 0)$.
(ii) One whose vertex sits exactly on the $x$-axis.
(iii) One that has NO real $x$-intercepts and whose $y$-intercept is at $(0, 10)$.
For each, write your chosen value of $c$, the full equation, the vertex, and a one-sentence reason that confirms the requirement is met.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $y = x^2 - 9$
Vertex: $(0, -9)$. $y$-intercept: $(0, -9)$ (same as vertex). $x$-intercepts: set $y = 0$: $x^2 = 9 \Rightarrow x = \pm 3$, so $(3, 0)$ and $(-3, 0)$. Range: $y \ge -9$.
1.2 — Vertex + real $x$-intercepts
(a) Vertex $(0, 1)$; $c = 1 > 0$, so NO real $x$-intercepts.
(b) Vertex $(0, 0)$; $c = 0$, so one $x$-intercept at $(0, 0)$ (the curve touches the axis at the vertex).
(c) Vertex $(0, -1)$; $c = -1 < 0$, so two real $x$-intercepts at $x = \pm 1$, i.e. $(1, 0)$ and $(-1, 0)$.
(d) Vertex $(0, -100)$; $c = -100 < 0$, so two real $x$-intercepts at $x = \pm 10$.
1.3 — Through $(2, 7)$
Use $y = x^2 + c$. Sub the point: $7 = (2)^2 + c = 4 + c$, so $c = 3$. Equation: $y = x^2 + 3$. Vertex: $(0, 3)$. $y$-intercept: $(0, 3)$.
1.4 — Three tables
$y = x^2$: $4, 1, 0, 1, 4$.
$y = x^2 + 3$: $7, 4, 3, 4, 7$ (each entry is $3$ more than the same column of $y = x^2$).
$y = x^2 - 2$: $2, -1, -2, -1, 2$ (each entry is $2$ less than the same column of $y = x^2$).
Every row is a vertical slide of the $y = x^2$ row by exactly $c$ — confirming all three curves have the same shape, just translated up or down.
1.5 — TRUE / FALSE
(a) TRUE — sub $x = 0$ into $y = x^2 + c$ gives $y = c$, and this is the lowest point.
(b) TRUE — $0 = x^2 - 4 \Rightarrow x = \pm 2$, two real roots.
(c) TRUE — sub $x = 0$ gives $y = 0 + 1 = 1$.
(d) FALSE — the $y$-intercept of $y = x^2 + c$ is $(0, c)$, which is negative whenever $c < 0$ (e.g. $y = x^2 - 4$ has $y$-intercept $(0, -4)$).
1.6 — $x$-intercepts at $\pm 5$
Sub $(5, 0)$: $0 = 5^2 + c = 25 + c$, so $c = \mathbf{-25}$. Check with $(-5, 0)$: $0 = (-5)^2 + c = 25 + c$ — same equation. Equation: $y = x^2 - 25$. Vertex: $\mathbf{(0, -25)}$.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student took only the positive square root, missing the negative one. The equation $x^2 = 25$ has TWO solutions: $x = 5$ and $x = -5$. Because $y = x^2 - 25$ is symmetric about the $y$-axis, $x$-intercepts always come in $\pm$ pairs.
(c) Corrected working:
Set $y = 0$: $0 = x^2 - 25$.
Rearrange: $x^2 = 25$.
Take both square roots: $x = \pm 5$.
So the $x$-intercepts are $\mathbf{(5, 0)}$ AND $\mathbf{(-5, 0)}$.
This is exactly the trap from the lesson's "Common Pitfalls" card — "dropping the negative root".
3 — Three designed parabolas (sample solution)
(i) $x$-intercepts at $(\pm 6, 0)$ require $0 = 36 + c$, so $c = \mathbf{-36}$. Equation: $y = x^2 - 36$. Vertex: $(0, -36)$. Reason: substituting $x = \pm 6$ gives $y = 36 - 36 = 0$ ✓.
(ii) Vertex on $x$-axis means the vertex has $y = 0$, so $c = \mathbf{0}$. Equation: $y = x^2$. Vertex: $(0, 0)$. Reason: the basic parabola sits with its vertex on the $x$-axis — this is exactly $y = x^2$.
(iii) $y$-intercept $(0, 10)$ means $c = \mathbf{10}$. Equation: $y = x^2 + 10$. Vertex: $(0, 10)$. Reason: $c = 10 > 0$ pushes the vertex above the $x$-axis, so the upward parabola never crosses it — no real $x$-intercepts. Also, sub $x = 0$ gives $y = 10$ ✓.
Marking: 1 mark for each correctly designed parabola with the right $c$ and reasoning, plus 1 mark for clear writing across all three.