Functions and graph transformations – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(2\times16-3=\) 29 (1) Square 4 first, then multiply by 2 and subtract 3. So \(f(4)=29\).
   2. \(2x^2=18\) (1), \(x^2=9\) (1), \(x=3\) or \(x=-3\) (1) Add 3 to both sides, divide by 2, then take the square root, remembering both signs. Check: \(2\times9-3=15\).

   In (a), a common error is to work out \((2\times4)^2-3=61\): the index applies only to \(x\), not to \(2x\). In (b), giving only \(x=3\) loses the final mark, since \((-3)^2=9\) as well.

2. Answer to Question 2 (4 marks)

   1. \(fg(x)=3x+4\) and \(gf(x)=3(x+4)=3x+12\) (1 each) For \(fg\), apply \(g\) first then \(f\). For \(gf\), apply \(f\) first then \(g\), and expand the bracket.
   2. \(3x+4=3x+12\) gives \(4=12\), which is never true, so there is no solution (1); the order of the functions always matters here (1) Subtracting \(3x\) from both sides leaves \(4=12\), which is false for every \(x\).

   For example, with \(x=1\), \(fg(1)=7\) but \(gf(1)=15\).

3. Answer to Question 3 (4 marks)

   1. \(y=\dfrac{x-5}2\) so \(x=2y+5\) (1): \(f^{-1}(x)=2x+5\) (1) Multiply both sides by 2, then add 5. Write the inverse using \(x\).
   2. \(f^{-1}(7)=19\) (1); \(f(19)=\dfrac{14}2=7\) ✓ (1) Substitute 7 into the inverse: \(2\times7+5=19\). Applying \(f\) to 19 gives back 7.

   In (a), another method is to reverse the steps of \(f\): \(f\) subtracts 5 then halves, so \(f^{-1}\) doubles then adds 5. A common error is to write \(f^{-1}(x)=\dfrac{2}{x-5}\), confusing the inverse with the reciprocal.

4. Answer to Question 4 (4 marks)

   1. \(f(x-2)\): \((5,6)\) (1); \(f(x)+4\): \((3,10)\) (1) \(f(x-2)\) moves the graph 2 units right, and \(+4\) moves it 4 units up.
   2. \((3,-6)\) (1), a minimum (1) \(-f(x)\) reflects the graph in the \(x\)-axis, so the maximum becomes a minimum.

   In (a), a common error is to move \(f(x-2)\) 2 units to the left; the minus sign inside the bracket moves the graph to the right. Only the \(x\)-coordinate changes for \(f(x-2)\), and only the \(y\)-coordinate changes for \(f(x)+4\).

5. Answer to Question 5 (4 marks)

   1. \((x-3)^2-9+5=\) \((x-3)^2-4\) (1 for \(a\), 1 for \(b\)) Halve \(-6\) to get \(-3\), then subtract \((-3)^2=9\) to keep the expression the same.
   2. Translation \(\begin{pmatrix}3\\-4\end{pmatrix}\) (1); minimum value \(-4\) (1) \(-3\) inside the bracket moves \(y=x^2\) 3 units right and \(-4\) moves it 4 units down. The minimum point is \((3,-4)\).

   Check (a) by expanding: \((x-3)^2-4=x^2-6x+9-4=x^2-6x+5\). The minimum occurs when \(x=3\).

Total : 20 marks