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IB Mathematics: Applications and Interpretation SL · 12.4: Solving Equations with Technology · 12.4.6

Where two graphs meet Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. The curve \(y = 3x^2 − 2x − 4\) and the line \(y = 2x + 5\) meet at two points. Find the coordinates of both points.

    • Graph \(y = 3x^2 − 2x − 4\) and \(y = 2x + 5\).
    • From the calculator: \(x = −1.19\) and \(x = 2.52\)
    • Use the line for \(y\): \(y = 2.62\) and \(y = 10.0\)
    • Points: \((−1.19, 2.62)\) and \((2.52, 10.0)\)
  2. Use your calculator to solve \(3^x = 3x^2\). Give every solution to 3 significant figures.

    • Graph \(y = 3^x\) and \(y = 3x^2\).
    • From the calculator: intersections at \(x = −0.451, 1.00, 3.00\)
    • The graphs cross three times, so there are three solutions.
  3. A model is \(g(x) = 4 + 15 × 0.7^x\).

    1. (a) Find the \(y\)-intercept of the graph of \(g\).

      • Put \(x = 0\): \(g(0) = 4 + 15 × 1 = 19\)
    2. (b) Write down the equation of the horizontal asymptote.

      • \(0.7^x\) gets closer to 0 as \(x\) grows, so \(g(x)\) gets closer to 4.
      • Asymptote: \(y = 4\)
    3. (c) Solve \(g(x) = 6\).

      • Graph \(y = 4 + 15 × 0.7^x\) and \(y = 6\).
      • From the calculator: \(x = 5.65\)

In context

  1. A tennis ball is hit upwards from a balcony. Its height above the ground after \(t\) seconds is \(h = −4.9t^2 + 9t + 6.2\) metres.

    1. (a) Find the greatest height and the time at which it is reached.

      • Graph \(y = −4.9x^2 + 9x + 6.2\) for \(x \geq 0\).
      • From the calculator: maximum at \(t = 0.918\), \(h = 10.3\)
    2. (b) Find when it hits the ground.

      • Set \(h = 0\): the positive root is \(t = 2.37\) s
      • The other root is negative, so it is rejected.
    3. (c) For how long is it higher than 9 m?

      • Intersections with \(y = 9\): \(t = 0.397\) and \(t = 1.44\)
      • Time above 9 m: \(1.44 − 0.397 = 1.04\) s
  2. A bakery sells \(x\) cakes a week. The weekly cost is \(C = 950 + 8x\) and the weekly revenue is \(R = 46x − 0.2x^2\), both in dollars.

    1. (a) Find the break-even quantities.

      • Graph \(y = 950 + 8x\) and \(y = 46x − 0.2x^2\).
      • From the calculator: \(x = 29.6\) and \(x = 160\)
    2. (b) Write down the smallest and the largest whole number of cakes that give a profit.

      • Profit when \(R > C\): between 29.6 and 160.
      • Whole numbers: 30 to 160
    3. (c) Find the greatest weekly profit, using a whole number of cakes.

      • Graph \(P = R − C = 38x − 0.2x^2 − 950\) and find the maximum.
      • Maximum near \(x = 95.0\); the whole number 95 gives \(P = 855.00\)
      • Greatest profit: $855.00
  3. The value of a laptop is \(V = 1800 × 0.78^t\) euros after \(t\) years. A student's savings are \(S = 300 + 420t\) euros, for \(0 \leq t \leq 6\).

    1. (a) Find when the two values are equal.

      • Graph \(y = 1800 × 0.78^x\) and \(y = 300 + 420x\) for \(0 \leq x \leq 6\).
      • From the calculator: one intersection, at \(t = 1.94\)
    2. (b) Which is greater after 4 years? Give both values.

      • At \(t = 4\): the laptop's value = €666 and the savings = €1 980
      • The savings are greater.
  4. A workshop makes \(n\) candles in one batch. The average cost of one candle, in euros, is \(A = 180/n + 0.5n + 4\).

    1. (a) Use your calculator to find the value of \(n\) that gives the lowest average cost, to 3 significant figures.

      • Graph \(y = 180/x + 0.5x + 4\) for \(x > 0\).
      • From the calculator: minimum at \(x = 19.0\), \(y = 23.0\)
    2. (b) The number of items must be a whole number. Find the best whole number and the average cost for it.

      • Compare the whole numbers either side: \(A(18) = 23.00\), \(A(19) = 22.97\)
      • The best whole number is 19, at €22.97 each.

Stretch

  1. The curve \(y = x^2 + 2x − 3\) and the line \(y = −x + 4\) meet at points A and B.

    −6−4−22−55101520xyAB
    1. (a) Find the coordinates of A and B.

      • Graph the curve and the line.
      • From the calculator: \(x = −4.54\) and \(x = 1.54\)
      • Use the line for \(y\): \(y = 8.54\) and \(y = 2.46\)
    2. (b) Find the length of AB.

      • \(AB = \sqrt{(1.54 − (−4.54))^2 + (2.46 − (8.54))^2}\)
      • \(AB = 8.60\)
  2. A rocket is launched from the ground. Its height is \(h_1 = −4.9t^2 + 25t\) metres after \(t\) seconds. At the same moment a balloon rises steadily from a height of 5 m, so its height is \(h_2 = 5 + 3t\) metres.

    1. (a) Find the times at which the two heights are equal.

      • Graph \(y = −4.9x^2 + 25x\) and \(y = 5 + 3x\)
      • From the calculator: intersections at \(t = 0.240\) and \(t = 4.25\)
    2. (b) For how long is the rocket higher than the balloon?

      • The rocket is higher between the two times.
      • \(4.25 − 0.240 = 4.01\) s

jmaths.xyz · AI SL · 12.4.6 Where two graphs meet