Year 10 Core — Sketching Parabolas using Transformations (7B)

Today's lesson

In 7A we met the basic parabola $y=x^2$ and saw how the number $a$ stretches or flips it. Today we put it all together: starting from $y=x^2$, we dilate, reflect and translate to sketch any parabola written in turning point form $y=a(x-h)^2+k$ — and we read its turning point straight off the rule.

Learning intentions

Name the three transformations: dilation, reflection and translation
Describe the effect of each on the graph of $y=x^2$
Read the turning point straight from turning point form $y=a(x-h)^2+k$
Sketch a parabola from turning point form, labelling the turning point and the $y$-intercept
Find the rule of a simple parabola given its turning point and one other point

Part 1 — The three transformations (~8 min)

Every parabola in this section is the basic $y=x^2$ that has been moved or reshaped. There are three things that can happen.

Dilation — the $a$ in $y=ax^2$ makes the curve narrower (large $|a|$) or wider (small $|a|$). The turning point stays at $(0,0)$.

Reflection — when $a<0$ the parabola is flipped in the $x$-axis: it becomes inverted (opens downward) and the turning point is now a maximum.

Translation — the brackets and the constant slide the curve without changing its shape. In $y=(x-h)^2$, the $h$ slides it sideways ($h>0$ → right, $h<0$ → left). In $y=x^2+k$, the $k$ slides it up or down ($k>0$ → up, $k<0$ → down).

Key idea — turning point form $y=a(x-h)^2+k$

The turning point is $(h,k)$.
The axis of symmetry is the vertical line $x=h$.
If $a>0$ the parabola is upright (minimum); if $a<0$ it is inverted (maximum).
The $y$-intercept is found by substituting $x=0$.

Watch the sign of $h$. The bracket is $(x-h)$, so $y=(x-3)^2$ has $h=3$ (turning point to the right) while $y=(x+5)^2$ means $h=-5$ (turning point to the left).

📺 Walkthrough: starting from $y=x^2$, how reflecting and then translating builds the graph of $y=-(x+1)^2+4$, with the turning point read straight from the rule.

Part 2 — Reading the turning point (Example 1, ~10 min)

The whole point of turning point form is that you can read $(h,k)$ without any working. Then substitute $x=0$ for the $y$-intercept.

EXAMPLE 1 — Turning point and y-intercept

For each rule, state the turning point (and whether it is a maximum or a minimum), the axis of symmetry, and the $y$-intercept. i $y=(x-3)^2-2$ ii $y=-(x+1)^2+4$.

i $y=(x-3)^2-2$: here $a=1>0$, $h=3$, $k=-2$, so it is a minimum at $(3,-2)$, axis $x=3$. $y$-intercept: $y=(0-3)^2-2=7$, so $(0,7)$.
ii $y=-(x+1)^2+4$: here $a=-1<0$, $h=-1$, $k=4$, so it is a maximum at $(-1,4)$, axis $x=-1$. $y$-intercept: $y=-(0+1)^2+4=3$, so $(0,3)$.

Now you try: State the turning point (max/min), axis and $y$-intercept of $y=(x+1)^2-2$ and $y=-(x-2)^2+3$. Answers: $y=(x+1)^2-2$ → min $(-1,-2)$, axis $x=-1$, $y$-int $(0,-1)$; $y=-(x-2)^2+3$ → max $(2,3)$, axis $x=2$, $y$-int $(0,-1)$.

Building understanding — write the turning point for each.

$y=x^2+3$
$y=-x^2-4$
$y=(x-2)^2$
$y=(x+5)^2$

a) $(0,3)$ b) $(0,-4)$ c) $(2,0)$ d) $(-5,0)$

Part 3 — Sketching from turning point form (Example 2, ~14 min)

To sketch: mark the turning point $(h,k)$, draw the axis of symmetry $x=h$, plot the $y$-intercept, then draw a smooth curve — opening up if $a>0$ and down if $a<0$.

EXAMPLE 2 — Sketching y=(x−3)²−2

Sketch $y=(x-3)^2-2$, labelling the turning point and the $y$-intercept.

$a=1>0$ (upright, minimum), turning point $(3,-2)$, axis $x=3$, $y$-intercept $(0,7)$.

$y=(x-3)^2-2$: minimum at $(3,-2)$, axis $x=3$, $y$-intercept $(0,7)$.

Now you try: Sketch $y=(x+1)^2-2$ — minimum $(-1,-2)$, axis $x=-1$, $y$-intercept $(0,-1)$.

EXAMPLE 3 — A reflected (inverted) parabola

Sketch $y=-(x+1)^2+4$, labelling the turning point and the $y$-intercept.

$a=-1<0$ (inverted, maximum), turning point $(-1,4)$, axis $x=-1$, $y$-intercept $(0,3)$. Because it opens downward, it also crosses the $x$-axis (here at $x=-3$ and $x=1$).

$y=-(x+1)^2+4$: inverted (maximum) at $(-1,4)$, axis $x=-1$, $y$-intercept $(0,3)$.

Now you try: Sketch $y=-(x-2)^2+3$ — maximum $(2,3)$, axis $x=2$, $y$-intercept $(0,-1)$.

Practice 3.1 — for each rule state: turning point, max or min, axis of symmetry, $y$-intercept.

$y=(x-1)^2+2$
$y=(x+4)^2-1$
$y=-(x-2)^2+5$
$y=-(x+3)^2$

a) $(1,2)$, min, $x=1$, $y$-int $(0,3)$
b) $(-4,-1)$, min, $x=-4$, $y$-int $(0,15)$
c) $(2,5)$, max, $x=2$, $y$-int $(0,1)$
d) $(-3,0)$, max, $x=-3$, $y$-int $(0,-9)$

Part 4 — Finding the rule from the graph (Example 4, ~8 min)

If a parabola has its turning point on the $y$-axis, its rule is $y=ax^2+k$. The $k$ is the turning-point height; substitute one other point to find $a$.

EXAMPLE 4 — Finding the rule

A parabola has turning point $(0,1)$ and passes through $(1,2)$. Find its rule.

Turning point on the $y$-axis, so $y=ax^2+1$ (here $k=1$). Substitute $(1,2)$:
$2=a(1)^2+1$, so $a=1$. The rule is $y=x^2+1$.

Now you try: Turning point $(0,-1)$ through $(1,1)$. Answer: $y=ax^2-1$, $1=a-1$ so $a=2$; rule $y=2x^2-1$.

Practice 4.1 — find each rule (each turning point is on the $y$-axis, so $y=ax^2+k$).

Turning point $(0,1)$, through $(2,-11)$.
Turning point $(0,10)$, through $(-3,7)$.

a) $y=ax^2+1$; $-11=4a+1$ so $a=-3$; rule $y=-3x^2+1$.
b) $y=ax^2+10$; $7=9a+10$ so $a=-\tfrac13$; rule $y=-\tfrac13 x^2+10$.

Part 5 — Quick quiz (5 min)

Working program — Cambridge Ex 7B (p600)

After the quiz, open Cambridge Chapter 7 (page 600) and complete the following:

Set work	Extension
Questions 1–6 (½), 8–9 (½), 10	Questions 7, 11 (½)

"(½)" means do every second part. Always mark the turning point, the axis of symmetry and the $y$-intercept on any sketch.

Exit ticket — write in your book

Before you pack up, write one sentence each:

For $y=a(x-h)^2+k$, what are the coordinates of the turning point?
How can you tell from the rule whether the turning point is a maximum or a minimum?
How do you find the $y$-intercept of any parabola from its rule?