Sketching using Factorisation (Ex 7C)

Year 10 Mathematics Core · Cambridge Ch 7, §7C · Mr Wong

Name

Class

Date

Learning intentions.

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

Key results

$y$-intercept: put $x=0$. $x$-intercepts: put $y=0$, factorise, then use the Null Factor Law ($p\times q=0 \Rightarrow p=0$ or $q=0$). The turning point sits halfway between the $x$-intercepts; substitute that $x$-value to get its $y$-coordinate.

Part A — Factorising for x-intercepts

1. Use the Null Factor Law to write both $x$-intercepts.

a $y=(x-1)(x+6)$

b $y=x(x+4)$

c $y=(x+3)(x-5)$

2. Factorise each, then state the $x$-intercept(s).

a $y=x^2-2x-3$

b $y=x^2-9$

c $y=x^2+4x+4$

d $y=x^2-6x$

Part B — Reading a sketch

3. The parabola below has rule $y=x^2-4x+3$. Use the graph to state each key feature.

a $y$-intercept

b $x$-intercepts

c Axis of symmetry

d Turning point

e Max or min?

Parabolas 7C — continued

Part C (factorise & sketch) · Part D (table) · Challenge

Name

Part C — Factorise, then sketch

4. For $y=x^2-2x-8$: factorise, find the $y$-intercept, both $x$-intercepts and the turning point, then sketch it on the grid.

a $y$-intercept

b $x$-intercepts

c axis of symmetry

d turning point

Part D — Table of values

5. Complete the table for $y=x^2-4x+3$, then state the turning point and the two $x$-intercepts.

$x$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y$

a turning point

b $x$-intercepts

Challenge

6. A parabola has $x$-intercepts $(3,0)$ and $(7,0)$ and turning point $(5,12)$. It has the form $y=m(x-3)(x-7)$.

a Why is the axis $x=5$?

b Substitute $(5,12)$ to find $m$.

c Write the full rule.