Solutions — Parabolas 7C Worksheet
Year 10 Mathematics Core · Cambridge Ch 7, §7C · Mr Wong
ANSWER KEY
Part A — Factorising for x-intercepts
Q1. Null Factor Law:
a $(1,0)$ and $(-6,0)$
b $(0,0)$ and $(-4,0)$
c $(-3,0)$ and $(5,0)$
Q2. Factorise, then $x$-intercept(s):
a $(x-3)(x+1)$: $x$-ints $3$ and $-1$.
b $(x-3)(x+3)$: $x$-ints $3$ and $-3$.
c $(x+2)^2$: one $x$-int at $-2$ (perfect square — touches).
d $x(x-6)$: $x$-ints $0$ and $6$.
Part B — Reading a sketch
Q3. The graphed parabola $y=x^2-4x+3$:
a $(0,3)$
b $(1,0)$ and $(3,0)$
c $x=2$
d $(2,-1)$
e minimum
Part C — Factorise, then sketch
Q4. $y=x^2-2x-8=(x-4)(x+2)$:
a $y$-intercept $(0,-8)$
b $x$-intercepts $(-2,0)$ and $(4,0)$
c axis of symmetry $x=1$
d turning point $(1,-9)$ (minimum)
Sketch: upright parabola through $(-2,0)$, $(4,0)$ and $(0,-8)$, with its lowest point at $(1,-9)$.
Part D — Table of values
Q5. Table for $y=x^2-4x+3$:
| $x$ | $-1$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|
| $y$ | 8 | 3 | 0 | -1 | 0 | 3 | 8 |
a turning point $(2,-1)$
b $x$-intercepts $(1,0)$ and $(3,0)$
Challenge
Q6. $x$-intercepts $(3,0)$, $(7,0)$, turning point $(5,12)$, form $y=m(x-3)(x-7)$:
a $x=5$ it is halfway between the intercepts: $\tfrac{3+7}{2}=5$.
b $m=-3$ $12=m(5-3)(5-7)=m(2)(-2)=-4m \Rightarrow m=-3$.
c $y=-3(x-3)(x-7)$ (opens downward, so $(5,12)$ is a maximum).