Solutions — Parabolas 7G Worksheet

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

ANSWER KEY

Warm-up

Q1. Vertical line meets parabola:

a $(2,12)$
b $(-1,-3)$

Q2. Rearranged:

a $x^2+3x+6=0$
b $x^2-5x+3=0$

Part A — Solving by substitution

Q3. Points of intersection:

a $(0,0)$ and $(3,9)$
b $(-1,-1)$ and $(2,2)$
c $(1,2)$ only (tangent)

Q4. $x^2=2x-1 \Rightarrow x^2-2x+1=0 \Rightarrow (x-1)^2=0$, so $x=1$, $y=1$.

· One point $(1,1)$ — the line is a tangent.

Part B — Working & sketching

Q5. $x^2-x-2=x+1$:

a $x^2-2x-3=0$
b $x=-1$ or $x=3$
c $(-1,0)$ and $(3,4)$

Part C — Using the quadratic formula

Q6. $x^2+2x-1=x+3$:

a $x^2+x-4=0$
b $(1.56,4.56)$ and $(-2.56,0.44)$

Part D — Discriminant

Q7. Number of intersection points:

a $\Delta=0$ → one (tangent)
b $\Delta=-15$ → none
c $\Delta=-3$ → none
d $\Delta=-24$ → none

Challenge

Q8. $x^2=x+k \Rightarrow x^2-x-k=0$. Tangent means $\Delta=0$: $(-1)^2-4(1)(-k)=1+4k=0$.

· $k=-\tfrac14$