Graphs and transformations – Year 12 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Each factor gives a root when it equals zero. The curve crosses the \(x\)-axis at \(x=-2\), \(x=1\) and \(x=3\). When \(x=0\): \(y=2\times(-1)\times(-3)=6\), so the \(y\)-intercept is \((0,6)\).
   2. The coefficient of \(x^3\) is positive, so \(y\to+\infty\) as \(x\to+\infty\) and \(y\to-\infty\) as \(x\to-\infty\). So the curve comes up from the bottom left and goes off to the top right.
   3. Read from the sketch where the curve lies below the \(x\)-axis. The curve is below the \(x\)-axis for \(x<-2\) and for \(1\lt x<3\). So the solution is \(x<-2\) or \(1\lt x<3\).

2. Answer to Question 2

   1. A squared factor gives a repeated root. The curve touches the \(x\)-axis at \((2,0)\) but does not cross it: there is a turning point on the axis. The curve changes direction there instead of crossing.
   2. Substitute \(x=0\). \(y=(-2)^2\times1=\) 4. So the \(y\)-intercept is \((0,4)\).
   3. \((x-2)^2\ge0\) always, so \(y\ge0\) when \(x+1\ge0\) or when \(x=2\): \(x\ge-1\) (and \(x=2\) is included in this).
   4. A translation by \(\begin{pmatrix}0\\-4\end{pmatrix}\) subtracts 4 from every \(y\)-value. New equation \(y=(x-2)^2(x+1)-4\). The \(y\)-intercept is \(4-4=\) 0, i.e. the curve passes through the origin.

3. Answer to Question 3

   1. The denominator is zero when \(x=2\), so one asymptote is \(x=2\). As \(x\to\pm\infty\), \(\dfrac3{x-2}\to0\) so \(y\to1\): the other asymptote is \(y=1\). The curve approaches these two lines but never meets them.
   2. Find the intercepts by setting \(x=0\) and then \(y=0\). Put \(x=0\): \(y=\dfrac3{-2}+1=-\dfrac12\). Point \((0,-\tfrac12)\). Put \(y=0\): \(\dfrac3{x-2}=-1\), \(x-2=-3\), \(x=-1\). Point \((-1,0)\). So \(C\) meets the axes at \((0,-\tfrac12)\) and \((-1,0)\).
   3. Compare \(y=\dfrac3{x-2}+1\) with \(y=\dfrac1x\): the 3 multiplies the output, the \(-2\) moves the graph right and the \(+1\) moves it up. First a stretch with scale factor 3 parallel to the \(y\)-axis (\(y=\tfrac3x\)), then a translation by \(\begin{pmatrix}2\\1\end{pmatrix}\). The asymptotes move from \(x=0\), \(y=0\) to \(x=2\), \(y=1\).

4. Answer to Question 4

   Changes outside the function act on the \(y\)-coordinate in the expected way; changes inside the function act on the \(x\)-coordinate in the opposite way.

   Curve	Image of \(A\)	Reason
   \(y=f(x)+2\)	\((4,8)\)	\(y\)-values increase by 2
   \(y=f(x-3)\)	\((7,6)\)	shift 3 right: \(x+3\)
   \(y=2f(x)\)	\((4,12)\)	\(y\)-values doubled
   \(y=f(2x)\)	\((2,6)\)	\(x\)-values halved
   \(y=-f(x)\)	\((4,-6)\)	reflection in the \(x\)-axis
   \(y=f(-x)\)	\((-4,6)\)	reflection in the \(y\)-axis

   For example, \(f(2x)\) halves the \(x\)-coordinate because \(2\times2=4\) gives the original input \(x=4\), and \(f(x-3)\) moves the point 3 to the right because \(7-3=4\).

5. Answer to Question 5

   1. A translation by \(\begin{pmatrix}2\\-1\end{pmatrix}\) replaces \(x\) by \(x-2\) and subtracts 1 from the whole function. Replace \(x\) by \((x-2)\) and subtract 1: \(y=(x-2)^2-3(x-2)-1=x^2-4x+4-3x+6-1=\) \(x^2-7x+9\).
   2. A reflection in the \(x\)-axis changes the sign of every \(y\)-value. Multiply \(y\) by \(-1\): \(y=-x^2+3x\).
   3. A stretch of scale factor \(\tfrac12\) parallel to the \(x\)-axis replaces \(x\) by \(2x\): \(y=(2x)^2-3(2x)=\) \(4x^2-6x\). Each \(x\)-coordinate of the image is half the original one.
   4. A reflection in the \(y\)-axis replaces \(x\) by \(-x\). Replace \(x\) by \(-x\): \(y=(-x)^2-3(-x)=\) \(x^2+3x\).

6. Answer to Question 6

   1. Inside the bracket, \(+3\) moves the graph left; outside, \(-4\) moves it down. A translation by the vector \(\begin{pmatrix}-3\\-4\end{pmatrix}\) (3 left, 4 down).
   2. Multiplying the whole function by 3 multiplies every \(y\)-coordinate by 3. A stretch with scale factor 3 parallel to the \(y\)-axis.
   3. Replacing \(x\) by \(\tfrac x2\) acts on the input. A stretch with scale factor 2 parallel to the \(x\)-axis (the \(x\)-coordinates double).
   4. Every \(y\)-coordinate changes sign. A reflection in the \(x\)-axis.
   5. Replacing \(x\) by \(4x\) divides each \(x\)-coordinate by 4. A stretch with scale factor \(\tfrac14\) parallel to the \(x\)-axis.

7. Answer to Question 7

   1. The lowest point is the vertex, midway between the towers at \(x=30\); use the top of the left tower, \((0,24)\), to find \(a\). Lowest point \((30,4)\), so \(y=a(x-30)^2+4\). At \(x=0\), \(y=24\): \(900a+4=24\), \(a=\dfrac{20}{900}=\dfrac1{45}\). So \(y=\dfrac1{45}(x-30)^2+4\).
   2. Start from \(y=x^2\), apply the vertical stretch, then move the vertex to \((30,4)\). A stretch with scale factor \(\tfrac1{45}\) parallel to the \(y\)-axis, then a translation by \(\begin{pmatrix}30\\4\end{pmatrix}\).
   3. 10 m from the left tower means \(x=10\) (by symmetry, \(x=50\) gives the same height). \(x=10\): \(y=\dfrac{(-20)^2}{45}+4=\dfrac{400}{45}+4=8.89+4=\) 12.89 m.
   4. Set \(y=10\) and solve for \(x\). \(\dfrac{(x-30)^2}{45}=6\), so \((x-30)^2=270\), \(x-30=\pm16.43\), giving \(x=13.57\) (or 46.43). About 13.57 m from the left tower. By symmetry, the rope is also 13.57 m from the right tower.

8. Answer to Question 8

   1. Substitute each speed into \(t=\dfrac{120}v\). 60 km/h: \(120\div60=\) 2 hours. 80 km/h: \(120\div80=\) 1.5 hours.
   2. Adding 0.5 to the output moves the graph up 0.5. A translation by \(\begin{pmatrix}0\\0.5\end{pmatrix}\) (up by 0.5). Original asymptotes: \(v=0\) and \(t=0\). New asymptotes: \(v=0\) and \(t=0.5\).
   3. Substitute \(v=100\) into the formula with the delay. \(\dfrac{120}{100}+0.5=\) 1.7 hours.
   4. Form an inequality; since \(v>0\) we can multiply through by \(v\) without changing the direction. \(\dfrac{120}v+0.5\le1.8\) so \(\dfrac{120}v\le1.3\), \(v\ge\dfrac{120}{1.3}=\) 92.3 km/h. The minimum average speed is 92.3 km/h.

9. Answer to Question 9

   1. Substitute each value of \(x\) into the formula. \(x=9\): \(36\div9=4\), \(4+3=7\) ✓. With the code: \(x=3\): \(10+3=13\); \(x=6\): \(5+3=8\); \(x=15\): \(2+3=5\).
   2. Compare \(\dfrac{36}x+3\) with \(\dfrac1x\). It is a stretch with scale factor 36 parallel to the \(y\)-axis followed by a translation by \(\begin{pmatrix}0\\3\end{pmatrix}\).
   3. Set the cost per person equal to 5 and solve. \(\dfrac{36}x+3=5\) gives \(\dfrac{36}x=2\), \(x=\) 18. 18 friends are needed without the code.
   4. Repeat with the cheaper order. \(\dfrac{30}x+3=5\) gives \(\dfrac{30}x=2\), \(x=\) 15. With the code, 15 friends are enough.

10. Answer to Question 10

    1. Substitute each value of \(x\) into \(h\). Each factor is zero in turn: \(x=0\), \(x=4\), \(x=9\).
    2. It is a cubic with positive leading coefficient: rises from the origin to a maximum between 0 and 4, falls below the axis between 4 and 9 and rises again after 9. At \(x=10\): \(h=\tfrac16\times10\times6\times1=10\).
    3. Use the sketch: the curve is above the axis between the first two roots and after the third. \(h>0\) for \(0\lt x<4\) and \(x>9\) (up to \(x=10\)).
    4. Substitute \(x=2\) and convert from tens of metres to metres. \(h(2)=\tfrac16\times2\times(-2)\times(-7)=\tfrac{28}6=4.67\) tens of metres, i.e. about 46.7 m.

11. Answer to Question 11

    1. Find a few points on each curve, such as the \(y\)-intercepts. Both curves are increasing and positive. \(y=3^x\) passes through \((0,1)\) and \((1,3)\). \(y=3^{x+1}\) passes through \((0,3)\) and \((-1,1)\). For \(x\le-2\) both are very close to the axis.
    2. Use the index law \(a^{m+n}=a^m\times a^n\). \(3^{x+1}=3^x\times3^1=3\times3^x\). So the second graph is a translation by \(\begin{pmatrix}-1\\0\end{pmatrix}\) or equivalently a stretch with scale factor 3 parallel to the \(y\)-axis.
    3. The asymptote is \(y=0\) for both.
    4. Take logarithms of both sides. \(3^{x+1}=100\): \(x+1=\dfrac{\log100}{\log3}=4.19\), so \(x=3.19\). After about 3.2 days.

12. Answer to Question 12

    1. Substitute the values into \(y=x^2(x-3)\). \(x=-1\): \(1\times(-4)=-4\) ✓. \(x=2\): \(4\times(-1)=-4\) ✓.
    2. Roots at \(x=0\) (touching, repeated) and \(x=3\). Local maximum at \((0,0)\) and local minimum at \((2,-4)\). The curve falls to \(-\infty\) on the left and rises on the right.
    3. \(k=-3\): 3 solutions. \(k=-4\): 2 distinct solutions (\(x=-1\) and the repeated \(x=2\)). \(k=0\): 2 solutions (\(x=0\) repeated and \(x=3\)). \(k=5\): 1 solution (the curve only reaches 5 once, for \(x>3\)).
    4. \((x+1)(x-2)^2=(x+1)(x^2-4x+4)=x^3-3x^2+4\) ✓. So \(x^2(x-3)=-4\) has the solution \(x=-1\) and a repeated solution \(x=2\) where the line touches the minimum.

13. Answer to Question 13

    1. Substitute the values into the formula. \(t=5\): \(20+60\times2^{-1}=20+30=50\) ✓. \(t=15\): \(20+60\times2^{-3}=20+7.5=27.5\) ✓.
    2. As \(t\to\infty\), \(2^{-t/5}\to0\). \(T=20\). The temperature approaches the room temperature (20°C) but never goes below it.
    3. Reflection in the \(y\)-axis (\(2^{-t}\)), then a stretch with scale factor 5 parallel to the \(t\)-axis (\(2^{-t/5}\)), then a stretch with scale factor 60 parallel to the \(T\)-axis, then a translation by \(\begin{pmatrix}0\\20\end{pmatrix}\).
    4. Solve the inequality with logarithms. \(20+60\times2^{-t/5}<40\) gives \(2^{-t/5}\lt \tfrac13\), so \(\dfrac t5>\log_23=1.585\), \(t>7.9\). After about 7.9 minutes. The drink is safe to hold after about 7.9 minutes.

14. Answer to Question 14

    1. Adding 3 to the function moves every point up 3. \(B\to(-1,7)\), \(C\to(1,1)\).
    2. Replacing \(x\) by \(x-2\) moves the graph 2 units right. Add 2 to each \(x\)-coordinate: \(A\to(-1,0)\), \(B\to(1,4)\), \(C\to(3,-2)\), \(D\to(5,0)\).
    3. Multiplying the function by 2 doubles every \(y\)-coordinate. Double the \(y\)-coordinates: \(B\to(-1,8)\), \(C\to(1,-4)\).
    4. Replacing \(x\) by \(2x\) halves every \(x\)-coordinate. Halve the \(x\)-coordinates: \(A\to(-\tfrac32,0)\) and \(D\to(\tfrac32,0)\).
    5. A reflection in the \(x\)-axis turns a maximum into a minimum and vice versa. Change the sign of \(y\): \(B\to(-1,-4)\) and \(C\to(1,2)\). The new maximum is \((1,2)\) (the image of \(C\)) and the new minimum is \((-1,-4)\).

15. Answer to Question 15

    1. Take out the common factor \(x\), then use the difference of two squares. \(x^3-4x=x(x^2-4)=x(x-2)(x+2)\). The curve crosses the axis at \(-2\), \(0\) and \(2\), rising on the right and falling on the left.
    2. Replacing \(x\) by \(x-1\) moves the graph 1 unit right. Translating 1 to the right moves the roots to \(-1\), \(1\) and \(3\): \(y=(x-1)(x-3)(x+1)\).
    3. The \(y\)-intercept of \(y=f(x-1)\) is the value at \(x=0\), which is \(f(-1)\). \(f(-1)=-1+4=\) 3, so \((0,3)\).
    4. Replace \(x\) by \(x+2\) in the factorised form. \(y=f(x+2)=(x+2)x(x+4)\). At \(x=0\): \(y=2\times0\times4=0\) ✓. Its roots are \(-4\), \(-2\), \(0\).

16. Answer to Question 16

    1. The denominator is zero at \(x=1\), and as \(x\to\pm\infty\) the fraction tends to 0 so \(y\to3\). \(x=1\) and \(y=3\). These are the vertical and horizontal asymptotes of \(C\).
    2. Set \(x=0\) for the \(y\)-intercept and \(y=0\) for the \(x\)-intercept. \(y\)-intercept: \(\dfrac2{-1}+3=1\), point \((0,1)\). \(x\)-intercept: \(\dfrac2{x-1}=-3\), \(x=1-\dfrac23=\dfrac13\), point \((\tfrac13,0)\). So \(C\) crosses the axes at \((0,1)\) and \((\tfrac13,0)\).
    3. Set the two expressions for \(y\) equal and multiply through by \(x-1\). \(\dfrac2{x-1}+3=x+1\Rightarrow\dfrac2{x-1}=x-2\Rightarrow2=(x-2)(x-1)=x^2-3x+2\Rightarrow x^2-3x=0\). This is the required equation.
    4. Factorise and substitute each root into the line \(y=x+1\). \(x(x-3)=0\) gives \(x=0\) and \(x=3\). Points \((0,1)\) and \((3,4)\). Check on \(C\): \(\dfrac2{3-1}+3=4\) ✓.

17. Answer to Question 17

    1. A quadratic with minimum \((p,q)\) can be written \(y=a(x-p)^2+q\); use the given point to find \(a\). Vertex form: \(y=a(x-3)^2-4\). At \((0,5)\): \(9a-4=5\), \(a=1\). So \(y=(x-3)^2-4=x^2-6x+5\). Check: \(0-0+5=5\) ✓.
    2. Touching the axis at \(x=-2\) gives a squared factor \((x+2)^2\); crossing at \(x=1\) gives a factor \((x-1)\). \(y=a(x+2)^2(x-1)\). At \(x=0\): \(a\times4\times(-1)=-8\), \(a=2\). So \(y=2(x+2)^2(x-1)\). Check: \(2\times4\times(-1)=-8\) ✓.
    3. Substitute \(x=2\) into \(y=2(x+2)^2(x-1)\). \(y=2\times16\times1=\) 32. The height of the cubic at \(x=2\) is 32.
    4. A reflection in the \(x\)-axis changes the sign of \(y\), so the minimum becomes a maximum. New equation \(y=-(x-3)^2+4\) (or \(y=-x^2+6x-5\)), maximum at \((3,4)\). The vertex \((3,-4)\) maps to \((3,4)\).

18. Answer to Question 18

    1. A reflection in the \(x\)-axis changes \(x^2\) into \(-x^2\); a translation upwards then adds 9. Reflect \(y=x^2\) in the \(x\)-axis: \(y=-x^2\). Translate by \(\begin{pmatrix}0\\9\end{pmatrix}\): \(y=9-x^2\).
    2. Substitute into \(y=9-x^2\). \(x=1\): \(9-1=\) 8 ft. \(x=2\): \(9-4=\) 5 ft.
    3. Replace \(x\) by \(\tfrac x2\): a stretch with scale factor 2 parallel to the \(x\)-axis. The new bottom edge: \(9-\tfrac{x^2}4=0\), \(x^2=36\), \(x=\pm6\), so the width is 12 ft (the old width was 6 ft).
    4. Set the two expressions equal. \(9-x^2=9-\tfrac{x^2}4\Rightarrow\tfrac34x^2=0\Rightarrow x=0\). They meet only at the top, \((0,9)\).

19. Answer to Question 19

    1. Take out \(x^2\) and use the difference of two squares. \(x^4-4x^2=x^2(x^2-4)=x^2(x-2)(x+2)\). The curve crosses the axis at \(x=\pm2\) and touches it at the origin. The \(x^4\) coefficient is positive, so both ends go up. There is a maximum at \((0,0)\) and minima at \((\pm\sqrt2,-4)\).
    2. \(f(-x)=(-x)^4-4(-x)^2=x^4-4x^2=f(x)\). The graph is symmetrical about the \(y\)-axis.
    3. \(x^2>0\) for \(x\ne0\), so \(f(x)<0\) when \((x-2)(x+2)<0\), i.e. \(-2\lt x<2\), \(x\ne0\).
    4. Adding 4 to \(f(x)\) moves the curve up 4 units. New equation \(y=x^4-4x^2+4=(x^2-2)^2\). It touches the \(x\)-axis where \(x^2=2\), i.e. at \((\sqrt2,0)\) and \((-\sqrt2,0)\).

20. Answer to Question 20

    1. The given curve is \(g(x)=x^2-6x+12\); undo the translation. The image is \(g(x)=f(x-2)-1\). So \(f(x-2)=g(x)+1=x^2-6x+13\). Replace \(x\) by \(x+2\): \(f(x)=(x+2)^2-6(x+2)+13=x^2+4x+4-6x-12+13=\) \(x^2-2x+5\). Check: \(f(x-2)-1=(x-2)^2-2(x-2)+5-1=x^2-6x+12\) ✓.
    2. A quadratic crosses the \(x\)-axis only if its discriminant \(b^2-4ac\) is at least 0. \(f\): discriminant \(4-20=-16<0\). \(g\): discriminant \(36-48=-12<0\). Neither has real roots, so neither crosses the axis.
    3. Two distinct points need a positive discriminant. \(f(x)-k=x^2-2x+5-k\) has discriminant \(4-4(5-k)=4k-16>0\) when \(k>4\). So the curve meets the axis twice when \(k>4\).
    4. Complete the square to find the turning point. \(f(x)=(x-1)^2+4\) has its minimum at \((1,4)\). Translate by \(\begin{pmatrix}0\\-4\end{pmatrix}\) to place it on the \(x\)-axis.