Mathematics • Year 9 • Unit 2 • Lesson 5
Vertical Translations: $y = x^2 + c$
Build the "shift by $c$, read off the vertex" routine. Work through one fully-worked example, then a faded guided example, then eight graduated independent problems — including intercept hunting.
1. I do — fully worked example
Read every line. The constant $c$ shifts everything vertically; the shape stays put.
Problem. For $y = x^2 - 4$: build a table for $x = -3, -2, -1, 0, 1, 2, 3$, state the vertex, and find both $x$-intercepts.
Step 1 — Build the $y = x^2$ row first.
$x^2$: $9, 4, 1, 0, 1, 4, 9$
Reason: $y = x^2 + c$ is built by taking the $y = x^2$ row and adding $c$ to every entry.
Step 2 — Add $c = -4$ to each entry.
$y = x^2 - 4$: $5, 0, -3, -4, -3, 0, 5$
Reason: subtracting $4$ from every $y$ shifts the whole curve down by $4$ units.
Step 3 — Read the vertex straight off.
Vertex is at $(0, c) = (0, -4)$.
Reason: vertex always sits at $(0, c)$. The lowest $y$-value in the table is $-4$, occurring at $x = 0$.
Step 4 — Find the $x$-intercepts: set $y = 0$.
$0 = x^2 - 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
Reason: rearrange and take BOTH square roots — the parabola is symmetric, so $x$-intercepts come in $\pm$ pairs.
Answer: Table $\mathbf{5, 0, -3, -4, -3, 0, 5}$. Vertex $\mathbf{(0, -4)}$. $x$-intercepts: $\mathbf{(2, 0)}$ and $\mathbf{(-2, 0)}$.
2. We do — fill in the missing steps
Same template as Section 1, with the working faded. 4 marks
Problem. For $y = x^2 + 3$: build a table for $x = -2, -1, 0, 1, 2$, state the vertex, and decide whether there are real $x$-intercepts.
Step 1 — Build the $y = x^2$ row:
$x^2$: $\_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_$
Step 2 — Add $c = 3$:
$y$: $\_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_$
Step 3 — Vertex: $(0, c) = (\_\_\_\_, \_\_\_\_)$.
Step 4 — $x$-intercepts: set $y = 0$: $0 = x^2 + 3 \Rightarrow x^2 = \_\_\_\_$ . Are there real solutions? __________________ . Reason in one sentence: __________________________________ .
3. You do — independent practice
Show your working. Foundation = read off the vertex. Standard = evaluate or find intercepts. Extension = construct an equation.
Foundation — read the vertex
3.1 State the vertex of $y = x^2 + 7$. 1 mark
3.2 State the vertex of $y = x^2 - 10$. 1 mark
3.3 Find $y$ when $x = 4$ on $y = x^2 - 5$. 1 mark
3.4 Find the $y$-intercept of $y = x^2 - 6$. 1 mark
Standard — evaluate and find intercepts
3.5 Find the $x$-intercepts of $y = x^2 - 16$. Show both roots. 2 marks
3.6 Explain in one or two sentences why $y = x^2 + 4$ has no real $x$-intercepts. 2 marks
Extension — construct the equation
3.7 Write the equation of a parabola with the same shape as $y = x^2$, opening upward, with vertex $(0, -4)$. Then find both $x$-intercepts. 3 marks
3.8 The parabola $y = x^2 + c$ passes through $(3, 5)$. Find $c$, write the equation, and state the vertex. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = x^2 + 3$)
Step 1: $x^2$: $\mathbf{4, 1, 0, 1, 4}$.
Step 2: $y$: $\mathbf{7, 4, 3, 4, 7}$.
Step 3: vertex $(\mathbf{0}, \mathbf{3})$.
Step 4: $x^2 = \mathbf{-3}$. No real solutions, because the square of any real number cannot be negative.
3.1 — Vertex of $y = x^2 + 7$
Read $c = 7$. Vertex: $\mathbf{(0, 7)}$.
3.2 — Vertex of $y = x^2 - 10$
Read $c = -10$. Vertex: $\mathbf{(0, -10)}$.
3.3 — $y = x^2 - 5$ at $x = 4$
$y = 4^2 - 5 = 16 - 5 = \mathbf{11}$.
3.4 — $y$-intercept of $y = x^2 - 6$
Sub $x = 0$: $y = 0^2 - 6 = -6$. $y$-intercept: $\mathbf{(0, -6)}$ (which is also the vertex).
3.5 — $x$-intercepts of $y = x^2 - 16$
Set $y = 0$: $0 = x^2 - 16 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$. $x$-intercepts: $\mathbf{(4, 0)}$ and $\mathbf{(-4, 0)}$.
3.6 — No real $x$-intercepts for $y = x^2 + 4$
Setting $y = 0$ gives $x^2 + 4 = 0 \Rightarrow x^2 = -4$. There is no real number whose square is negative, so the curve does not cross the $x$-axis. Geometrically the vertex $(0, 4)$ sits above the $x$-axis, and the upward parabola never comes back down to it.
3.7 — Vertex $(0, -4)$ equation and intercepts
Same shape as $y = x^2$ means we use $y = x^2 + c$. Vertex $(0, c) = (0, -4)$, so $c = -4$. Equation: $\mathbf{y = x^2 - 4}$.
$x$-intercepts: set $y = 0$: $x^2 - 4 = 0 \Rightarrow x = \pm 2$. So $\mathbf{(2, 0)}$ and $\mathbf{(-2, 0)}$.
3.8 — Parabola through $(3, 5)$
Sub the point: $5 = (3)^2 + c = 9 + c$, so $c = \mathbf{-4}$. Equation: $\mathbf{y = x^2 - 4}$. Vertex: $\mathbf{(0, -4)}$.