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IB Mathematics: Applications and Interpretation HL · 12.5: Functions · 12.5.7

Composite transformations Problem set B Worked solutions

Skills

  1. Write down the equation of the graph after each change.

    1. (a) The graph of \(y=2x^5\) is stretched vertically with scale factor 7.

      • Multiply the whole function by 7: \(y=7\times2x^5\)
      • \(y=14 x^{5}\)

    2. (b) The graph of \(y=5 x^{3} + 2 x^{2}\) is reflected in the x-axis and then reflected in the y-axis.

      • Reflect in the x-axis: \(y=-(5 x^{3} + 2 x^{2})=- 5 x^{3} - 2 x^{2}\)
      • Reflect in the y-axis: replace x by −x
      • \(y=5 x^{3} - 2 x^{2}\)

    3. (c) The graph of \(y=2^x\) is translated 3 units in the negative x direction and then stretched vertically with scale factor 6.

      • Translate: replace x by \(x+3\): \(y=2^{x+3}\)
      • Stretch: multiply by 6
      • \(y=6\times 2^{x+3}\)

    4. (d) The graph of \(y=x^{2} + 3 x\) is translated 2 units in the positive x direction and then stretched vertically with scale factor 3. Give your answer in the form \(y=ax^2+bx+c\).

      • Translate: \(y=(x-2)^2+3(x-2)\)
      • Stretch: multiply everything by 3
      • \(y=3 x^{2} - 3 x - 6\)

  2. A vertical stretch can sometimes be replaced by a horizontal stretch.

    1. (a) The graph of \(y=x^2\) is stretched vertically with scale factor 25. Write down the equation of the new graph.

      • Multiply the function by 25: \(y=25x^2\)

    2. (b) Find the scale factor of the horizontal stretch that has the same effect on \(y=x^2\).

      • \((qx)^2=q^2x^2\), so \(q^2=25\) and \(q=5\)
      • Horizontal stretch with scale factor \(1/q=0.2\)

    3. (c) The graph of \(y=3x^3\) is stretched vertically with scale factor 64. Find the scale factor of the horizontal stretch that has the same effect.

      • \(3(qx)^3=3q^3x^3\), so \(q^3=64\) and \(q=4\)
      • Horizontal stretch with scale factor \(1/q=0.25\)

  3. The diagram shows the graph of \(y=f(x)\), a cubic. On separate diagrams sketch each graph, marking the turning points.

    −3−2−1123−8−6−4−2246xyy = f(x)(−1, 3)(1, −5)

    1. (a) \(y=2f(x-2)\)

      • Turning points of f: (−1, 3) and (1, −5)
      • Translate 2 right: add 2 to each x-coordinate
      • Stretch vertically with scale factor 2: multiply each y-coordinate by 2
      • Image points: (1, 6) and (3, −10)

    2. (b) \(y=\tfrac12f(x+3)\)

      • Turning points of f: (−1, 3) and (1, −5)
      • Translate 3 left: subtract 3 from each x-coordinate
      • Stretch vertically with scale factor 0.5: multiply each y-coordinate by 0.5
      • Image points: (−4, 1.5) and (−2, −2.5)

  4. The diagram shows the graph of \(y=f(x)\), a cubic. On separate diagrams sketch each graph, marking the turning points.

    −3−2−1123−8−6−4−2246xyy = f(x)(−1, 3)(1, −5)

    1. (a) \(y=-f(x)+4\)

      • Turning points of f: (−1, 3) and (1, −5)
      • Stretch/reflect first: multiply each y-coordinate by −1
      • Then translate: add 4 to each y-coordinate
      • Image points: (−1, 1) and (1, 9)

    2. (b) \(y=2-3f(x)\)

      • Turning points of f: (−1, 3) and (1, −5)
      • Stretch/reflect first: multiply each y-coordinate by −3
      • Then translate: add 2 to each y-coordinate
      • Image points: (−1, −7) and (1, 17)

  5. The diagram shows the graph of \(y=f(x)\), a cubic. On separate diagrams sketch each graph, marking the turning points.

    −3−2−1123−8−6−4−2246xyy = f(x)(−1, 3)(1, −5)

    1. (a) \(y=f(\tfrac14x+3)\)

      • Turning points of f: (−1, 3) and (1, −5)
      • Translate by −3 in the x direction: subtract 3 from each x-coordinate
      • Then stretch horizontally with scale factor 4: multiply each x-coordinate by 4
      • Image points: (−16, 3) and (−8, −5)

    2. (b) \(y=f(2x-4)\)

      • Turning points of f: (−1, 3) and (1, −5)
      • Translate by 4 in the x direction: add 4 to each x-coordinate
      • Then stretch horizontally with scale factor 0.5: multiply each x-coordinate by 0.5
      • Image points: (1.5, 3) and (2.5, −5)

  6. The graph of \(y=f(x)\) is made of straight lines joining (−6, 0), (−3, 4), (0, 4), (3, −2), (6, 0). Sketch each graph.

    −8−6−4−22468−4−2246xy

    1. (a) \(y=3f(x+1)-2\)

      • Horizontal: subtract 1 from each x-coordinate
      • Vertical: multiply each y-coordinate by 3, then subtract 2 from each y-coordinate
      • Image points: (−6, 0)→(−7, −2), (−3, 4)→(−4, 10), (0, 4)→(−1, 10), (3, −2)→(2, −8), (6, 0)→(5, −2)
      • Join the points with straight lines

    2. (b) \(y=f(3x+6)\)

      • Horizontal: subtract 6 from each x-coordinate, then multiply each x-coordinate by 1/3
      • No vertical change
      • Image points: (−6, 0)→(−4, 0), (−3, 4)→(−3, 4), (0, 4)→(−2, 4), (3, −2)→(−1, −2), (6, 0)→(0, 0)
      • Join the points with straight lines

  7. Describe a sequence of transformations that maps the graph of \(y=3 x^{2} + x\) to the graph of the given equation.

    1. (a) \(y=12 x^{2} + 4 x - 5\)

      • Let \(f(x)=3 x^{2} + x\)
      • \(12 x^{2} + 4 x - 5=4(3 x^{2} + x)-5=4f(x)-5\)
      • The stretch comes before the translation

    2. (b) \(y=3 x^{2} + 13 x + 11\)

      • \(y=f(x+2)-3\)
      • Both changes are translations, so the order does not matter

  8. The graph of \(f(x)=0.5^x\) has the horizontal asymptote \(y=0\). For each change write the equation of the image and the equation of its horizontal asymptote.

    1. (a) A translation 5 units down.

      • Translate down 5: \(y=0.5^x-5\)
      • The asymptote moves down 5: \(y=-5\)

    2. (b) A reflection in the x-axis, then a translation 2 units up.

      • Reflect in the x-axis: \(y=-0.5^x\), asymptote \(y=0\)
      • Translate up 2: \(y=-0.5^x+2\), asymptote \(y=2\)

    3. (c) A vertical stretch with scale factor 5, then a translation 3 units to the right and 1 unit up.

      • Stretch: \(y=5\times0.5^x\), asymptote \(y=0\)
      • Translate 3 right and 1 up: \(y=5\times0.5^{x-3}+1\), asymptote \(y=1\)

  9. The graph of \(y=\dfrac{1}{x}\) has the asymptotes \(x=0\) and \(y=0\). The graph is translated 4 units to the right and 1 unit up. Write the equation of the image and the equations of its asymptotes.

    • Translate 4 right: replace x by \(x-4\): \(y=\dfrac{1}{x-4}\)
    • Translate 1 up: \(y=\dfrac{1}{x-4}+1\)
    • The vertical asymptote moves 4 right: \(x=4\). The horizontal asymptote moves 1 up: \(y=1\).

In context

  1. The cross-section of a skateboard ramp is modelled by \(y=\frac{x^{2}}{4} - 2 x + 6\), where x and y are in metres.

    1. (a) Write \(\frac{x^{2}}{4} - 2 x + 6\) in the form \(a(x-h)^2+k\).

      • \(\frac{x^{2}}{4} - 2 x + 6=\frac{1}{4}(x^2-8x)+6\)
      • \(=\frac{1}{4}\big[(x-4)^2-16\big]+6\)
      • \(=\frac{1}{4}(x-4)^2+2\)

    2. (b) Describe two transformations that map the graph of \(y=x^2\) to this curve.

      • With \(f(x)=x^2\): \(y=\frac{1}{4}f(x-4)+2\)
      • Vertical stretch with scale factor 0.25, then translate 4 in the x direction and 2 in the y direction

  2. A graph shows the fare F(d) in euros of a taxi journey against the distance d. A new tariff gives the fare \(N=1.5F+12\) euros.

    1. (a) Describe the transformations that map the graph of F to the graph of N.

      • N\(=1.5\times\)(old value)\(+12\)
      • A multiplication comes first, then the addition: stretch, then translate

    2. (b) The same graph can be made by translating first and stretching second. Describe that translation and that stretch.

      • \(1.5y+12=1.5\big(y+8\big)\)
      • So translate up 8 first, then stretch with scale factor 1.5

Stretch

  1. Describe the transformations that map \(y=\sqrt{x}\) to each graph, in order.

    1. (a) \(y=\sqrt{4x+2}\)

      • \(\sqrt{4x+2}=g(4x)\) where \(g(x)=\sqrt{x+2}\)
      • First \(g\): translate by −2. Then replace x by 4x: stretch with scale factor \(1/4\)

    2. (b) \(y=\sqrt{4(x+2)}\)

      • \(\sqrt{4(x+2)}=h(x+2)\) where \(h(x)=\sqrt{4x}\)
      • First \(h\): stretch with scale factor \(1/4\). Then translate by −2

    3. (c) Where does the point \((0,0)\) of \(y=\sqrt{x}\) end up in each graph?

      • First graph: x = −2 before the stretch, then −2 ÷ 4 = −0.5
      • Second graph: the stretch leaves 0 where it is, then the translation gives −2
      • The two graphs are different: the order matters

  2. Find two different sequences of two transformations that map \(y=f(x)\) to \(y=f(4x-10)\).

    • As written: \(f(x-10)\) first, then replace x by 4x
    • Sequence 1: translate 10, then stretch with scale factor \(1/4\)
    • Factor: \(4x-10=4\big(x-2.5\big)\)
    • Sequence 2: stretch with scale factor \(1/4\), then translate 2.5

  3. The graph of \(y=ax^2+bx+c\) is translated 2 units in the positive x direction and then stretched horizontally with scale factor \(\tfrac13\). The resulting graph is \(y=9x^2+cx-3\). Find a, b and c.

    • The new graph is \(y=f(3x-2)\) with \(f(x)=ax^2+bx+c\)
    • \(a(3x-2)^2+b(3x-2)+c=9 a x^{2} + 4 a - 2 b + c + x \left(- 12 a + 3 b\right)\)
    • Compare: \(9 a=9\) gives a = 1; the x term gives \(- 12 a + 3 b=c\); the constant gives \(4 a - 2 b + c=-3\)
    • Solving these three equations: a = 1, b = 5, c = 3

  4. The graph of \(y=f(x)\) is stretched vertically with scale factor 4 away from the line \(y=-3\) (the line \(y=-3\) does not move).

    1. (a) Write down the equation of the new graph in terms of f.

      • Translate so the line is the x-axis: \(y=f(x)+3\)
      • Stretch: \(y=4\big(f(x)+3\big)\)
      • Translate back: \(y=4\big(f(x)+3\big)-3=4f(x)+9\)

    2. (b) The point (2, 5) is on the graph of \(y=f(x)\). Find its image.

      • The x-coordinate stays 2
      • y: \(4\times5+9=29\)

jmaths.xyz · AI HL · 12.5.7 Composite transformations