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

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1. Answer to Question 1

   1. \(f(4)=12-2=\) 10; \(f(-1)=-3-2=\) \(-5\); \(f(0.5)=1.5-2=\) \(-0.5\). Replace \(x\) by each value: multiply by 3, then subtract 2.
   2. \(3x-2=19\), so \(3x=21\) and \(x=7\). Set the expression equal to 19, add 2 to both sides, then divide by 3. Check: \(f(7)=21-2=19\).
   3. \(g(3)=9+1=\) 10; \(g(-2)=4+1=\) 5. Square first, then add 1; a negative number squared is positive.
   4. \(x^2+1=26\), so \(x^2=25\) and \(x=5\) or \(x=-5\). Subtract 1 and take the square root, remembering both signs. Check: \(25+1=26\).

2. Answer to Question 2

   1. \(fg(x)=f(x^2)=2x^2+3\). \(gf(x)=g(2x+3)=(2x+3)^2=4x^2+12x+9\).
   2. \(fg(2)=2\times4+3=\) 11. \(gf(2)=7^2=\) 49. Substitute \(x=2\) into the simplified expressions from part (a).
   3. \(2x^2+3=35\), so \(x^2=16\) and \(x=4\) or \(x=-4\). Subtract 3 from both sides and divide by 2, then take the square root, remembering \(\pm\).
   4. The order matters: \(fg\) means "do \(g\) first, then \(f\)", which is a different order of operations from \(gf\).

3. Answer to Question 3

   1. Let \(y=4x-7\). Then \(y+7=4x\) and \(x=\dfrac{y+7}4\). So \(f^{-1}(x)=\dfrac{x+7}4\). Undo the operations in reverse order: add 7, then divide by 4. Write the answer using \(x\).
   2. \(y=\dfrac{x+2}5\) gives \(5y=x+2\), so \(g^{-1}(x)=5x-2\). Multiply both sides by 5, then subtract 2.
   3. \(y=\dfrac x3+1\) gives \(y-1=\dfrac x3\), so \(h^{-1}(x)=3(x-1)=3x-3\). Subtract 1, then multiply by 3.
   4. \(f^{-1}(9)=\dfrac{16}4=\) 4. Check: \(f(4)=16-7=9\). ✓ A function followed by its inverse always returns the starting value.

4. Answer to Question 4

   1. The smallest value is \(f(0)=0\) and the largest is \(f(-3)=9\). The range is \(0\le f(x)\le9\).
   2. \(x=0\) gives division by zero, so \(x\ne0\).
   3. You cannot take the square root of a negative number, so \(x-1\ge0\): domain \(x\ge1\). The output is always at least 0: range \(h(x)\ge0\).
   4. When \(x=0\), \(k=1\), and \(k\) increases, so the range is \(k(x)\ge1\).

5. Answer to Question 5

   1. Up 2: \((3,6)\).
   2. \(f(x+1)\) moves the graph 1 left: \((2,4)\).
   3. Stretch vertically by 2: \((3,8)\).
   4. Horizontal stretch by scale factor \(\tfrac12\): \((1.5,4)\).
   5. Reflection in the \(x\)-axis: \((3,-4)\).
   6. Reflection in the \(y\)-axis: \((-3,4)\).

6. Answer to Question 6

   1. Roots: \((x-1)(x-3)=0\), so it crosses the \(x\)-axis at \(x=1\) and \(x=3\). The \(y\)-intercept is 3. Completing the square: \((x-2)^2-1\), so the minimum (turning point) is \((2,-1)\). It is a U-shaped parabola.
   2. Passes through \((0,1)\), \((1,2)\), \((2,4)\), \((3,8)\) and \((-1,0.5)\). It rises steeply on the right and approaches the \(x\)-axis (\(y=0\)) on the left but never touches it.
   3. Two curved branches, one in the top-right quadrant and one in the bottom-left. It gets close to both axes but never meets them (asymptotes).

7. Answer to Question 7

   1. \(2.5+1.6\times5=\) £10.50. Substitute \(d=5\) into the formula: \(1.6\times5=8\).
   2. \(y=2.5+1.6d\), so \(d=\dfrac{y-2.5}{1.6}\): \(f^{-1}(x)=\dfrac{x-2.5}{1.6}\). It gives the distance for a given fare.
   3. \(\dfrac{15.70-2.50}{1.6}=\dfrac{13.2}{1.6}=\) 8.25 miles. Substitute \(x=15.70\) into the inverse function.
   4. \(gf(d)=1.1(2.5+1.6d)=2.75+1.76d\). For \(d=5\): \(2.75+8.8=\) £11.55. Apply \(f\) first, then \(g\), and expand the bracket.

8. Answer to Question 8

   1. \(gh(x)=1.15(x-3)\): the euros received for \(x\) pounds, after the fee. Apply \(h\) first (take off the fee), then \(g\) (convert to euros).
   2. \(1.15\times197=\) 226.55 euros. Substitute \(x=200\): \(200-3=197\), then multiply by 1.15.
   3. \(y=1.15(x-3)\), so \(\dfrac{y}{1.15}=x-3\) and \((gh)^{-1}(x)=\dfrac x{1.15}+3\). Divide both sides by 1.15, then add 3.
   4. \(\dfrac{345}{1.15}+3=300+3=\) £303. So you need £303 to receive 345 euros.

9. Answer to Question 9

   1. \(\dfrac59\times45=\) 25°C. Subtract 32 first: \(77-32=45\), then multiply by \(\tfrac59\). So 77 °F is 25 °C.
   2. \(y=\dfrac59(F-32)\), so \(\dfrac{9y}5=F-32\) and \(C^{-1}(x)=\dfrac{9x}5+32\). Multiply both sides by \(\tfrac95\), then add 32. Write the answer using \(x\).
   3. \(\dfrac{9\times37}5+32=66.6+32=\) 98.6°F. Substitute \(x=37\) into the inverse: \(9\times37=333\) and \(333\div5=66.6\). Normal body temperature, 37 °C, is 98.6 °F.
   4. \(C(-40)=\dfrac59(-72)=-40\). ✓ \(-40-32=-72\) and \(\tfrac59\times(-72)=-40\). So \(-40\) °F is the same as \(-40\) °C.

10. Answer to Question 10

    1. \(gf(x)=1.2\times0.9x=\) \(1.08x\). Apply \(f\) first (the discount), then \(g\) (the VAT), and multiply the two multipliers.
    2. \(1.08\times250=\) £270. The laptop costs £270.
    3. \(fg(x)=0.9\times1.2x=1.08x=gf(x)\). Multiplication can be done in either order, so it does not matter.
    4. \(y=1.08x\), so \((gf)^{-1}(x)=\dfrac x{1.08}\). \(\dfrac{162}{1.08}=\) £150. Dividing by 1.08 undoes multiplying by 1.08.

11. Answer to Question 11

    1. \(h(0)=\) 20, \(h(1)=20-5=\) 15, \(h(2)=20-20=\) 0.
    2. The height goes from 20 down to 0: \(0\le h(t)\le20\).
    3. \(20-5t^2=10\), so \(5t^2=10\), \(t^2=2\) and \(t=1.41\) s.
    4. At \(t=2\) the ball reaches the ground (\(h=0\)), so the formula no longer applies.

12. Answer to Question 12

    1. \(f(x+4)\) moves the graph 4 units left: \((-2,-3)\).
    2. Down 5: \((2,-8)\).
    3. Stretch vertically by 2: \((2,-6)\).
    4. Horizontal stretch scale factor \(\tfrac12\): \((1,-3)\).
    5. A reflection in the \(x\)-axis turns the minimum into a maximum at \((2,3)\).

13. Answer to Question 13

    1. \(0.5,\ 1,\ 2,\ 4,\ 8\).
    2. The asymptote moves from \(y=0\) up to \(y=3\). At \(x=0\): \(1+3=\) 4.
    3. A translation of 1 unit to the left (vector \(\begin{pmatrix}-1\\0\end{pmatrix}\)).
    4. \(2^{x+1}=2^x\times2^1=2\times2^x\), so shifting left by 1 gives the same graph as stretching vertically by scale factor 2.

14. Answer to Question 14

    1. \(15+8m=30+5m\), so \(3m=15\) and \(m=5\) months. Set the two costs equal, then subtract \(5m\) and 15 from both sides.
    2. \(15+40=\) £55 on both. Substitute \(m=5\) into either function.
    3. \(y=15+8m\) gives \(m=\dfrac{y-15}8\): \(a^{-1}(x)=\dfrac{x-15}8\). \(\dfrac{111-15}8=\) 12 months. Subtract 15 and divide by 8.
    4. \(a(12)=15+96=111\) and \(b(12)=30+60=90\). Plan B is cheaper by £21.

15. Answer to Question 15

    1. \(fg(x)=3(x^2+2)-1=3x^2+5\). \(gf(x)=(3x-1)^2+2=9x^2-6x+1+2=9x^2-6x+3\). For \(fg\), put \(g(x)\) into \(f\); for \(gf\), put \(f(x)\) into \(g\).
    2. \(3x^2+5=53\), so \(3x^2=48\) and \(x^2=16\): \(x=4\) or \(x=-4\). Use \(fg(x)=3x^2+5\) from part (a).
    3. \(3x^2+5=9x^2-6x+3\), so \(0=6x^2-6x-2\), which divides by 2 to give \(3x^2-3x-1=0\). Set the two expressions equal and collect all the terms on one side.
    4. \(x=\dfrac{3\pm\sqrt{9+12}}6=\dfrac{3\pm\sqrt{21}}6\), so \(x=1.26\) or \(x=-0.26\). Use the quadratic formula with \(a=3\), \(b=-3\), \(c=-1\).

16. Answer to Question 16

    1. \(f(4)=\dfrac{9}{1}=\) 9. Substitute \(x=4\): the top is \(2\times4+1=9\) and the bottom is \(4-3=1\).
    2. Let \(y=\dfrac{2x+1}{x-3}\). Then \(y(x-3)=2x+1\), so \(xy-3y=2x+1\). Collect \(x\): \(xy-2x=1+3y\), so \(x(y-2)=1+3y\) and \(x=\dfrac{3y+1}{y-2}\). Swap the letters: \(f^{-1}(x)=\dfrac{3x+1}{x-2}\).
    3. \(f^{-1}(9)=\dfrac{28}7=\) 4. ✓ Substitute \(x=9\) into the inverse. It gives back 4, which matches part (a).
    4. \(x=2\) would give division by zero, so \(x\ne2\).

17. Answer to Question 17

    1. \(g(x)=(x+2)^2-5\), with turning point \((-2,-5)\). Replace \(x\) by \(x+2\) in \(x^2\), then subtract 5.
    2. A translation by the vector \(\begin{pmatrix}-2\\-5\end{pmatrix}\). \(+2\) inside the bracket moves the graph 2 left, and \(-5\) outside moves it 5 down.
    3. \((x+2)^2=5\), so \(x+2=\pm\sqrt5\) and \(x=-2\pm2.236\): \(x=0.24\) and \(x=-4.24\).
    4. The lowest value is \(-5\): \(g(x)\ge-5\). A square is never negative, so the smallest value of \(g(x)\) is \(-5\).

18. Answer to Question 18

    1. \(f(g(x))=g(x)+5=3x+2\), so \(g(x)=3x-3\).
    2. \(gf(x)=g(x+5)=3(x+5)-3=\) \(3x+12\). Replace \(x\) by \(x+5\) in \(g\).
    3. \(y=3x-3\), so \(x=\dfrac{y+3}3\): \(g^{-1}(x)=\dfrac{x+3}3\). Add 3 and divide by 3.
    4. \(3x-3=\dfrac{x+3}3\), so \(9x-9=x+3\) and \(8x=12\): \(x=1.5\). Check: \(g(1.5)=1.5\) and \(g^{-1}(1.5)=1.5\). ✓

19. Answer to Question 19

    1. \(f(x)+2\): \((0,2),(2,6),(4,2)\). \(f(x-1)\): \((1,0),(3,4),(5,0)\). \(-f(x)\): \((0,0),(2,-4),(4,0)\). \(f(2x)\): \((0,0),(1,4),(2,0)\). \(2f(x)\): \((0,0),(2,8),(4,0)\).
    2. Maximum height 4: the translation \(f(x-1)\) and the squash \(f(2x)\) both keep the height at 4.
    3. \(y=f(2x)\), whose points are \((0,0)\), \((1,4)\) and \((2,0)\).

20. Answer to Question 20

    1. \(ff(1)=f(f(1))=f(5)=17\). So \(a+b=5\) and \(5a+b=17\).
    2. Subtract: \(4a=12\), so \(a=3\) and \(b=2\). Check: \(f(x)=3x+2\), \(f(1)=5\), \(f(5)=17\). ✓
    3. \(y=3x+2\), so \(f^{-1}(x)=\dfrac{x-2}3\). Subtract 2 and divide by 3.
    4. \(3x+2=\dfrac{x-2}3\), so \(9x+6=x-2\) and \(8x=-8\): \(x=-1\). Check: \(f(-1)=-1\) and \(f^{-1}(-1)=-1\). ✓