Mathematics • Year 9 • Unit 2 • Lesson 7
Sideways Slides — Mixed Challenge
Pull every horizontal-translation skill together: find vertices, $y$-intercepts, axis of symmetry, and reverse the rule. Catch one classic mistake, then invent your own pair of parabolas to compare.
1. Mixed problems
Each problem mixes the skills of Lesson 7. Show working. 3 marks each
1.1 State the vertex and $y$-intercept of each parabola: (a) $y = (x - 10)^2$ (b) $y = (x + 1)^2$ (c) $y = (x - \tfrac{1}{2})^2$ (d) $y = x^2$ (think of this as $y = (x - 0)^2$).
1.2 $y = x^2$ is shifted: (a) 5 units RIGHT; (b) 3 units LEFT; (c) 0.5 units RIGHT. Write the new equation in each case and state the vertex.
1.3 Sketch $y = (x - 1)^2$ on a labelled set of axes. Mark the vertex, the $y$-intercept, and one extra symmetric pair (e.g. at $x = 3$ and $x = -1$). Show how the symmetric pair has matching $y$-values.
1.4 A parabola has the same shape as $y = x^2$ and passes through both $(2, 0)$ and $(0, 4)$. (a) Use $(2, 0)$ to deduce the vertex's $x$-coordinate. (b) Hence write the equation. (c) Check by substituting $(0, 4)$ into your equation.
1.5 For each statement, decide whether it is TRUE or FALSE. If false, write the corrected statement. (a) "The graph of $y = (x + 4)^2$ has axis of symmetry $x = 4$." (b) "The vertex of $y = (x - 6)^2$ is on the $x$-axis." (c) "The $y$-intercept of $y = (x - 3)^2$ is $(0, -9)$." (d) "$y = (x + 2)^2$ and $y = (x - 2)^2$ are reflections of each other across the $y$-axis."
1.6 Where do $y = (x - 2)^2$ and $y = (x + 4)^2$ intersect? (a) Set them equal: $(x - 2)^2 = (x + 4)^2$. Expand both sides. (b) Solve for $x$. (c) Substitute back to find the $y$-value at the intersection.
2. Find the mistake
A classmate has tried to read features off five horizontally-translated parabolas. Exactly two of their answers are wrong. Spot them, explain why, and fix them. 3 marks
Student's answers:
A: Vertex of $y = (x - 3)^2$ is $(-3, 0)$. (Negative because of the minus sign.)
B: Vertex of $y = (x + 5)^2$ is $(-5, 0)$. ✓
C: $y$-intercept of $y = (x - 4)^2$ is $(0, 16)$. ✓
D: $y = (x - 2)^2$ equals $x^2 - 4$. (Square both terms separately.)
E: Axis of symmetry of $y = (x + 1)^2$ is $x = -1$. ✓
(a) Which two are wrong?
(b) For each wrong one, explain in one sentence why the student's reasoning is mistaken.
(c) Write out the correct version of each wrong answer.
Stuck on D? $(x - 2)^2 = (x - 2)(x - 2) = x^2 - 4x + 4$. Not $x^2 - 4$.3. Open-ended challenge — design a pair of parabolas
This question has many valid answers. 4 marks
3.1 Design TWO parabolas of the form $y = (x - h)^2$ that satisfy ALL these conditions:
• One has its vertex on the POSITIVE $x$-axis.
• The other has its vertex on the NEGATIVE $x$-axis.
• The two parabolas share the same $y$-intercept value.
For your pair:
(i) Write both equations.
(ii) State both vertices.
(iii) Show that both have the same $y$-intercept.
(iv) Briefly explain WHY they end up with the same $y$-intercept (think about what $h^2$ does to a negative $h$).
Bonus: Find the $x$-value where the two parabolas intersect each other (set them equal and solve).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Vertex and $y$-intercept
(a) Vertex $(10, 0)$, $y$-int $(0, 100)$.
(b) Vertex $(-1, 0)$, $y$-int $(0, 1)$.
(c) Vertex $(\tfrac{1}{2}, 0)$, $y$-int $(0, \tfrac{1}{4})$.
(d) Vertex $(0, 0)$, $y$-int $(0, 0)$ (same point — this is the unshifted parabola).
1.2 — Shifts of $y = x^2$
(a) $h = +5$: $y = (x - 5)^2$, vertex $(5, 0)$.
(b) $h = -3$: $y = (x + 3)^2$, vertex $(-3, 0)$.
(c) $h = +0.5$: $y = (x - 0.5)^2$, vertex $(0.5, 0)$.
1.3 — Sketch $y = (x - 1)^2$
Vertex $(1, 0)$ (bracket zero at $x = 1$). $y$-intercept: $y = (0 - 1)^2 = 1$, so $(0, 1)$. Symmetric pair: at $x = 3$, $y = 4$; at $x = -1$, $y = 4$ (both 2 units from axis $x = 1$). Smooth upward U through five labelled points.
1.4 — Same shape, passes through $(2, 0)$ and $(0, 4)$
(a) $(2, 0)$ has $y = 0$, which only happens at the vertex of $y = (x - h)^2$, so the vertex is at $(2, 0)$. Hence $h = 2$. (b) Equation: $y = (x - 2)^2$. (c) Check $(0, 4)$: $y = (0 - 2)^2 = 4$. ✓ Yes — works.
1.5 — True / False
(a) FALSE. Bracket zero at $x = -4$, so axis is $x = -4$.
(b) TRUE. Vertex is $(6, 0)$, which lies on the $x$-axis.
(c) FALSE. Sub $x = 0$: $y = (-3)^2 = 9$, not $-9$. $y$-intercept is $(0, 9)$.
(d) TRUE. $y = (x + 2)^2$ has vertex $(-2, 0)$; $y = (x - 2)^2$ has vertex $(2, 0)$. Both have the same shape — the second is the mirror of the first across the $y$-axis.
1.6 — Intersection of $(x - 2)^2$ and $(x + 4)^2$
(a) $(x - 2)^2 = x^2 - 4x + 4$; $(x + 4)^2 = x^2 + 8x + 16$. (b) Set equal: $x^2 - 4x + 4 = x^2 + 8x + 16 \Rightarrow -4x + 4 = 8x + 16 \Rightarrow -12 = 12x \Rightarrow x = -1$. (c) At $x = -1$: $y = (-1 - 2)^2 = 9$ (and $y = (-1 + 4)^2 = 9$ — both check out). Intersection point: $(-1, 9)$. Note: $-1$ is the midpoint of the two vertices $x = 2$ and $x = -4$ — that's where two equal-shape parabolas always meet.
2 — Find the mistake
(a) The two wrong answers are A and D.
(b) A: The student followed the sign of the minus inside the bracket. But the vertex is where the bracket = 0, which is $x = +3$ for $(x - 3)$, not $x = -3$. D: The student treated $(x - 2)^2$ as if you could square each term separately. You can't — squaring a binomial gives three terms (FOIL), not two. $(x - 2)^2 = x^2 - 4x + 4$, not $x^2 - 4$.
(c) A corrected: Vertex of $y = (x - 3)^2$ is $(3, 0)$. D corrected: $(x - 2)^2 = x^2 - 4x + 4$.
Both mistakes are flagged in the lesson — "follow-the-sign" trap and "squaring before subtracting" pitfall.
3 — Open-ended challenge (sample solutions)
Pair example: $y_1 = (x - 3)^2$ and $y_2 = (x + 3)^2$.
(ii) Vertices: $(3, 0)$ (positive $x$-axis) and $(-3, 0)$ (negative $x$-axis).
(iii) $y$-intercept of $y_1$: $(0 - 3)^2 = 9$. $y$-intercept of $y_2$: $(0 + 3)^2 = 9$. Both are $(0, 9)$.
(iv) The $y$-intercept of $y = (x - h)^2$ is $h^2$. Squaring kills the sign: $(-3)^2$ and $(3)^2$ both equal $9$. So a positive $h$ and a negative $h$ with the same absolute value produce the same $y$-intercept.
Bonus: Set $(x - 3)^2 = (x + 3)^2 \Rightarrow x^2 - 6x + 9 = x^2 + 6x + 9 \Rightarrow -12x = 0 \Rightarrow x = 0$. They intersect at $(0, 9)$ — exactly the shared $y$-intercept.
Marking: 1 mark for valid pair with vertices in correct halves; 1 mark for $y$-intercept calculation; 1 mark for the $h^2$ explanation; 1 mark for the intersection bonus or clear write-up.