Solutions — Parabolas 7D Worksheet

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

ANSWER KEY

Warm-up

Q1. $y=x^2+2x-5 = x^2+2x+1-1-5 = (x+1)^2-6$; TP $=(-1,-6)$.

Q2. Solve:

a $x=\pm3$

b $x=\pm\sqrt3$

c $x=5$ or $x=-3$

d $x=-4\pm\sqrt2$

Q3. Turning point form and TP:

a $(x-3)^2+1$; TP $(3,1)$

b $(x+2)^2-3$; TP $(-2,-3)$

c $(x-4)^2+4$; TP $(4,4)$

d $(x+5)^2-7$; TP $(-5,-7)$

Q4. $y=-3(x-2)^2+12$:

a maximum at $(2,12)$ ($a=-3<0$)

b $y$-intercept $(0,0)$ ($-3(0-2)^2+12=0$)

c $x$-intercepts $(0,0)$ and $(4,0)$ ($(x-2)^2=4\Rightarrow x=0,4$)

Q5. $y=x^2+6x+15 = (x+3)^2+6$.

a $(x+3)^2+6$

b TP $(-3,6)$ minimum

c $y$-int $(0,15)$

d no $x$-intercepts

Q6. $y=x^2-4x+2 = (x-2)^2-2$.

a $(x-2)^2-2$

b TP $(2,-2)$ minimum

c $y$-int $(0,2)$

d $x$-int $x=2\pm\sqrt2$

Q7. $x$-intercepts (exact):

a $x=1\pm\sqrt5$

b none ($(x+1)^2+4=0$)

c $x=3\pm\sqrt5$

d $x=\tfrac{3\pm\sqrt{13}}{2}$

Q8. TP at $(3,-4)$ means $y=(x-3)^2-4 = x^2-6x+5$.

a $p=-6,\ q=5$

b $x$-intercepts: $(x-3)^2=4 \Rightarrow$ $(1,0)$ and $(5,0)$