2019 H2 Mathematics Paper 1 Question 3

Graphs and Transformations

Answers

(i)

2{(x−1)3−5}{2\{(x-1)^3-5\}}

(ii)

  • Translate the graph of y=x3{y=x^3} by 1{1} unit in the positive x-{x\textrm{-}}axis direction.
  • Translate the resulting graph by 5{5} unit in the negative y-{y\textrm{-}}axis direction.
  • Scale the resulting graph by scale factor 2{2} parallel to the y-{y\textrm{-}}axis.

Full solutions

(i)

p{(x+q)3+r}=p(x3+3qx2+3q2x+q3+r)\begin{align*} & p\{ (x+q)^3 + r \} \\ & = p ( x^3 + 3qx^2 + 3q^2x + q^3 + r ) \end{align*}
2x3−6x2+6x−12=px3+3pqx2+3pq2x+p(q3+r)2x^3-6x^2+6x-12 = p x^3 + 3pqx^2 + 3pq^2x + p(q^3 + r )
Comparing coefficients,
x3:p=2x2:3pq=−66q=−6q=−1x0:p(q3+r)=−122(−1+r)=−12−1+r=−6r=−5\begin{align*} &x^3:& p = 2 \\ &x^2:& 3pq = -6 \\ && 6q &= -6 \\ && q &= -1 \\ &x^0:& p(q^3+r) &= -12 \\ && 2(-1+r) &= -12 \\ && -1+r &= -6 \\ && r &= -5 \end{align*}
We check the coefficient of x{x} to ensure the validity of the comparison:
RHS=3pq2=3(2)(−1)2=6=LHS\begin{align*} \textrm{RHS} &= 3pq^2\\ &= 3(2)(-1)^2 \\ &= 6 \\ &= \textrm{LHS} \end{align*}
f(x)=2{(x−1)3−5}  ■f(x)=2\{(x-1)^3-5\} \; \blacksquare

(ii)

  • Translate the graph of y=x3{y=x^3} by 1{1} unit in the positive x-{x\textrm{-}}axis direction.
  • Translate the resulting graph by 5{5} unit in the negative y-{y\textrm{-}}axis direction.
  • Scale the resulting graph by scale factor 2{2} parallel to the y-{y\textrm{-}}axis.