Mathematics • Year 9 • Unit 2 • Lesson 7
Horizontal Translations $y = (x - h)^2$
Build the "minus means right" reflex: zero the bracket to find the vertex, sub $x = 0$ for the $y$-intercept, and remember the shape is identical to $y = x^2$ — it has just slid sideways. 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. For $y = (x - 4)^2$: state the vertex, the axis of symmetry, the $y$-intercept, and describe how the curve has shifted from $y = x^2$.
Step 1 — Zero the bracket.
Set $x - 4 = 0 \Rightarrow x = 4$. So $h = 4$.
Reason: the smallest possible value of $(x - 4)^2$ is $0$, which happens exactly when the bracket is zero.
Step 2 — Vertex and axis.
Vertex $(h, 0) = (4, 0)$. Axis of symmetry $x = 4$.
Reason: every $y = (x - h)^2$ has its vertex sitting on the $x$-axis at $x = h$, and the axis is the vertical line through it.
Step 3 — $y$-intercept.
Sub $x = 0$: $y = (0 - 4)^2 = (-4)^2 = 16$. So $(0, 16)$.
Reason: $y$-intercept is always where the curve crosses the $y$-axis; substitute $x = 0$ to find $y$.
Step 4 — Describe the shift.
"Minus inside" means the curve has slid 4 units to the RIGHT compared to $y = x^2$.
Reason: the vertex moved from $(0, 0)$ to $(4, 0)$ — same shape, just relocated.
Answer: Vertex $(4, 0)$, axis $x = 4$, $y$-intercept $(0, 16)$, shifted 4 RIGHT.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill each blank. 4 marks
Problem. For $y = (x + 5)^2$: state the vertex, the axis of symmetry, the $y$-intercept, and describe the shift.
Step 1 — Rewrite in $(x - h)$ form: $(x + 5)^2 = (x - $__________$)^2$. So $h = $ __________ .
Step 2 — Vertex and axis: Vertex $(h, 0) = ($__________ , __________$)$. Axis of symmetry $x = $ __________ .
Step 3 — $y$-intercept: Sub $x = 0$: $y = (0 + 5)^2 = $__________ . So $(0, $__________ $)$.
Step 4 — Describe the shift: "Plus inside" $\Rightarrow$ the curve has shifted __________ units to the __________________ (LEFT / RIGHT).
3. You do — independent practice
Show working under each problem. Foundation: zero the bracket. Standard: vertex + $y$-intercept. Extension: reverse-engineer the equation from a feature.
Foundation — find the vertex
3.1 State the vertex of $y = (x - 7)^2$. 1 mark
3.2 State the vertex of $y = (x + 4)^2$. 1 mark
3.3 State the axis of symmetry of $y = (x - 2)^2$. 1 mark
3.4 Find the $y$-intercept of $y = (x - 6)^2$. Show working. 1 mark
Standard — vertex AND $y$-intercept together
3.5 For $y = (x - 2)^2$, state: (a) the vertex; (b) the axis of symmetry; (c) the $y$-intercept (show working for (c)). 2 marks
3.6 Sketch $y = (x + 3)^2$ on a labelled set of axes. Mark the vertex, the $y$-intercept, and one symmetric point at $x = -1$. 2 marks
Extension — reverse the rule
3.7 $y = x^2$ is shifted 8 units to the LEFT. (a) Write the new equation. (b) State the vertex. (c) Find the $y$-intercept. 2 marks
3.8 A parabola has the same shape as $y = x^2$ but its vertex is at $(-7, 0)$. (a) Write its equation. (b) Find its $y$-intercept. (c) Explain in one sentence why the sign inside the bracket is OPPOSITE to the sign of the vertex's $x$-coordinate. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = (x + 5)^2$)
Step 1: $(x + 5)^2 = (x - (\mathbf{-5}))^2$, so $h = \mathbf{-5}$.
Step 2: Vertex $(\mathbf{-5}, \mathbf{0})$. Axis $x = \mathbf{-5}$.
Step 3: $y = (0 + 5)^2 = \mathbf{25}$, so $(0, \mathbf{25})$.
Step 4: Curve shifted 5 units to the LEFT.
3.1 — Vertex of $y = (x - 7)^2$
Bracket zero at $x = 7$. Vertex $(7, 0)$.
3.2 — Vertex of $y = (x + 4)^2$
$(x + 4) = 0 \Rightarrow x = -4$. Vertex $(-4, 0)$.
3.3 — Axis of symmetry of $y = (x - 2)^2$
Vertex at $x = 2$, so axis is the vertical line $x = 2$.
3.4 — $y$-intercept of $y = (x - 6)^2$
Sub $x = 0$: $y = (0 - 6)^2 = (-6)^2 = 36$. So $(0, 36)$.
3.5 — $y = (x - 2)^2$
(a) Bracket zero at $x = 2$. Vertex $(2, 0)$. (b) Axis $x = 2$. (c) Sub $x = 0$: $y = (0 - 2)^2 = 4$. $y$-intercept $(0, 4)$.
3.6 — Sketch $y = (x + 3)^2$
Vertex $(-3, 0)$ (bracket zero at $x = -3$). $y$-intercept: $y = (0 + 3)^2 = 9$, so $(0, 9)$. Symmetric point at $x = -1$: $y = (-1 + 3)^2 = 4$, so $(-1, 4)$. Mirror across axis $x = -3$: $(-5, 4)$. Smooth upward U through all four labelled points.
3.7 — Shift left by 8
(a) Left by 8 means $h = -8$. Equation: $y = (x - (-8))^2 = (x + 8)^2$. (b) Vertex $(-8, 0)$. (c) $y$-intercept: $y = (0 + 8)^2 = 64$, so $(0, 64)$.
3.8 — Vertex $(-7, 0)$
(a) $y = (x + 7)^2 = (x - (-7))^2$. (b) Sub $x = 0$: $y = (0 + 7)^2 = 49$. $y$-intercept $(0, 49)$. (c) The vertex sits at the $x$ that makes the bracket zero. For the bracket to be zero at $x = -7$, the inside must be "$x + 7$" (which is "$x - (-7)$"). So the sign you SEE inside the bracket is the opposite of the sign of the vertex $x$-coordinate.