Year 10 Core — Sketching using Factorisation (7C)

Today's lesson

So far we have read a parabola straight from turning point form. Today we tackle the form $y=x^2+bx+c$ — where the turning point is hidden. The trick is to factorise to find the $x$-intercepts, then use the symmetry of the parabola to locate the turning point exactly halfway between them.

Learning intentions

Know the steps for sketching a graph of the form $y=x^2+bx+c$
Understand that a parabola can have 0, 1 or 2 $x$-intercepts
Use factorisation and the Null Factor Law to find the $x$-intercepts
Use symmetry (the midpoint of the $x$-intercepts) to find the turning point

Part 1 — The four steps (~8 min)

Key idea — sketching $y=x^2+bx+c$

$y$-intercept: substitute $x=0$ (you simply read off $c$).
$x$-intercepts: substitute $y=0$, factorise, then use the Null Factor Law — if $p\times q=0$ then $p=0$ or $q=0$.
Axis of symmetry is halfway between the $x$-intercepts; this is also the $x$-coordinate of the turning point.
Turning point: substitute that $x$-value back into the rule to find the $y$-coordinate.

Null Factor Law. A product is zero only when one of its factors is zero. So $0=(x-1)(x-5)$ means $x-1=0$ or $x-5=0$, giving $x=1$ or $x=5$.

Number of $x$-intercepts. A parabola can cross the $x$-axis twice (two factors), once (a perfect square, e.g. $(x+3)^2$) or not at all (it cannot be factorised over the real numbers).

📺 Walkthrough: factorising $y=x^2-6x+5$ to find the $x$-intercepts, then using symmetry to land the turning point and sketch the graph.

Part 2 — Finding the turning point from the x-intercepts (Example 1, ~14 min)

Once you know both $x$-intercepts, the turning point is easy: its $x$-coordinate is their average, and you substitute to get the $y$-coordinate.

EXAMPLE 1 — Factorise, then sketch

Sketch $y=x^2-6x+5$, showing the $y$-intercept, both $x$-intercepts and the turning point.

$y$-intercept: $x=0$ gives $y=5$, so $(0,5)$.
$x$-intercepts: $0=x^2-6x+5=(x-1)(x-5)$, so $x=1$ or $x=5$ → $(1,0)$ and $(5,0)$.
Axis of symmetry: $x=\dfrac{1+5}{2}=3$.
Turning point: $y=3^2-6(3)+5=-4$, so $(3,-4)$, a minimum.

$y=x^2-6x+5$: $x$-intercepts $1$ and $5$, axis $x=3$, turning point $(3,-4)$, $y$-intercept $(0,5)$.

Now you try: Sketch $y=x^2-8x+7$. Answer: $(x-1)(x-7)$, $x$-intercepts $1$ and $7$; axis $x=4$; turning point $(4,-9)$; $y$-intercept $(0,7)$.

EXAMPLE 2 — Factor of x (one intercept at the origin)

Find the $x$-intercepts and turning point of $y=x^2-2x$.

$0=x^2-2x=x(x-2)$, so $x=0$ or $x=2$ → $x$-intercepts $(0,0)$ and $(2,0)$.
Axis of symmetry $x=\dfrac{0+2}{2}=1$. Turning point: $y=1^2-2(1)=-1$, so $(1,-1)$, a minimum.

Now you try: $y=x^2-6x$. Answer: $x(x-6)$, $x$-intercepts $0$ and $6$; axis $x=3$; turning point $(3,-9)$.

Building understanding — use the Null Factor Law to write both $x$-intercepts.

$y=(x+1)(x-2)$
$y=x(x-3)$
$y=x^2-4x \;=\; x(x-4)$
$y=x^2+2x-8 \;=\; (x+4)(x-2)$

a) $(-1,0)$ and $(2,0)$ b) $(0,0)$ and $(3,0)$ c) $(0,0)$ and $(4,0)$ d) $(-4,0)$ and $(2,0)$

Part 3 — The perfect square (Example 3, ~8 min)

When a quadratic factorises to a perfect square, the two intercepts coincide — the parabola just touches the $x$-axis at its turning point.

EXAMPLE 3 — Sketching a perfect square

Sketch $y=x^2+6x+9$.

$y$-intercept: $x=0$ gives $y=9$, so $(0,9)$.
$x$-intercepts: $0=x^2+6x+9=(x+3)^2$, so $x+3=0$, i.e. $x=-3$ (one intercept only).
Turning point: the curve touches the $x$-axis there, so the turning point is $(-3,0)$, a minimum.

Now you try: Sketch $y=x^2+8x+16$. Answer: $(x+4)^2$, touches at $x=-4$; turning point $(-4,0)$; $y$-intercept $(0,16)$.

Part 4 — Finding the rule and a downward parabola (Example 4, ~10 min)

If you are given the $x$-intercepts, you can write the rule in factor form, then find the turning point by symmetry — and the same idea works when $a$ is negative and the parabola opens downward.

EXAMPLE 4 — A downward parabola

For $y=-x^2-2x+8$, find the $x$-intercepts and the turning point, then sketch.

Factor out $-1$: $y=-(x^2+2x-8)=-(x+4)(x-2)$. So $x=-4$ or $x=2$ → $x$-intercepts $(-4,0)$ and $(2,0)$.
Axis of symmetry $x=\dfrac{-4+2}{2}=-1$. Turning point: $y=-(-1)^2-2(-1)+8=9$, so $(-1,9)$.
Since $a=-1<0$ the parabola is inverted, so this turning point is a maximum. $y$-intercept: $(0,8)$.

$y=-x^2-2x+8$: inverted, $x$-intercepts $-4$ and $2$, axis $x=-1$, maximum $(-1,9)$, $y$-intercept $(0,8)$.

Now you try (finding the rule): A parabola has $x$-intercepts $-3$ and $1$ and rule of the form $y=(x+a)(x+b)$. Find the rule and the turning point. Answer: $y=(x+3)(x-1)$; axis $x=-1$; turning point $(-1,-4)$.

Practice 4.1 — factorise, then state both $x$-intercepts and the turning point.

$y=x^2-8x+12$
$y=x^2-25$
$y=x^2-8x+16$

a) $(x-2)(x-6)$; $x$-ints $2,6$; axis $x=4$; turning point $(4,-4)$.
b) $(x-5)(x+5)$; $x$-ints $-5,5$; axis $x=0$; turning point $(0,-25)$.
c) $(x-4)^2$; touches at $x=4$; turning point $(4,0)$.

Part 5 — Quick quiz (5 min)

Working program — Cambridge Ex 7C (p608)

After the quiz, open Cambridge Chapter 7 (page 608) and complete the following:

Set work	Extension
Questions 1–6 (½), 7, 8, 11, 12	Questions 9, 10, 13, 14

"(½)" means do every second part. For every sketch, show the $y$-intercept, both $x$-intercepts (from factorising) and the turning point.

Exit ticket — write in your book

Before you pack up, write one sentence each:

How do you find the $x$-intercepts of $y=x^2+bx+c$?
Once you know both $x$-intercepts, how do you find the axis of symmetry?
How many $x$-intercepts does a perfect-square quadratic such as $(x+3)^2$ have, and why?