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

Where two graphs meet Problem set A Worked solutions

Skills

  1. The curve \(y = x^2 − 5x + 3\) and the line \(y = 2x − 6\) meet at two points. Find the coordinates of both points.

    • Graph \(y = x^2 − 5x + 3\) and \(y = 2x − 6\).
    • From the calculator: \(x = 1.70\) and \(x = 5.30\)
    • Use the line for \(y\): \(y = −2.61\) and \(y = 4.61\)
    • Points: \((1.70, −2.61)\) and \((5.30, 4.61)\)
  2. Use your calculator to solve \(2^x = x^2\). Give every solution to 3 significant figures.

    • Graph \(y = 2^x\) and \(y = x^2\).
    • From the calculator: intersections at \(x = −0.767, 2.00, 4.00\)
    • The graphs cross three times, so there are three solutions.
  3. A model is \(g(x) = 5 + 12 × 0.8^x\).

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

      • Put \(x = 0\): \(g(0) = 5 + 12 × 1 = 17\)
    2. (b) Write down the equation of the horizontal asymptote.

      • \(0.8^x\) gets closer to 0 as \(x\) grows, so \(g(x)\) gets closer to 5.
      • Asymptote: \(y = 5\)
    3. (c) Solve \(g(x) = 7\).

      • Graph \(y = 5 + 12 × 0.8^x\) and \(y = 7\).
      • From the calculator: \(x = 8.03\)

In context

  1. A stone is thrown upwards from a bridge. Its height above the river after \(t\) seconds is \(h = −4.9t^2 + 12t + 9.5\) metres.

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

      • Graph \(y = −4.9x^2 + 12x + 9.5\) for \(x \geq 0\).
      • From the calculator: maximum at \(t = 1.22\), \(h = 16.8\)
    2. (b) Find when it hits the river.

      • Set \(h = 0\): the positive root is \(t = 3.08\) s
      • The other root is negative, so it is rejected.
    3. (c) For how long is it higher than 12 m?

      • Intersections with \(y = 12\): \(t = 0.230\) and \(t = 2.22\)
      • Time above 12 m: \(2.22 − 0.230 = 1.99\) s
  2. A workshop sells \(x\) toys a week. The weekly cost is \(C = 1500 + 12x\) and the weekly revenue is \(R = 60x − 0.25x^2\), both in dollars.

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

      • Graph \(y = 1500 + 12x\) and \(y = 60x − 0.25x^2\).
      • From the calculator: \(x = 39.3\) and \(x = 153\)
    2. (b) Write down the smallest and the largest whole number of toys that give a profit.

      • Profit when \(R > C\): between 39.3 and 153.
      • Whole numbers: 40 to 152
    3. (c) Find the greatest weekly profit, using a whole number of toys.

      • Graph \(P = R − C = 48x − 0.25x^2 − 1500\) and find the maximum.
      • Maximum near \(x = 96.0\); the whole number 96 gives \(P = 804.00\)
      • Greatest profit: $804.00
  3. The value of a car is \(V = 24 000 × 0.85^t\) dollars after \(t\) years. The amount still owed on its loan is \(L = 21 000 − 1500t\) dollars, for \(0 \leq t \leq 8\).

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

      • Graph \(y = 24000 × 0.85^x\) and \(y = 21000 − 1500x\) for \(0 \leq x \leq 8\).
      • From the calculator: one intersection, at \(t = 1.54\)
    2. (b) Which is greater after 5 years? Give both values.

      • At \(t = 5\): the car's value = $10 649 and the loan = $13 500
      • The loan is greater.
  4. A print shop makes \(n\) posters in one batch. The average cost of one poster, in dollars, is \(A = 250/n + 0.8n + 3\).

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

      • Graph \(y = 250/x + 0.8x + 3\) for \(x > 0\).
      • From the calculator: minimum at \(x = 17.7\), \(y = 31.3\)
    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: \(A(17) = 31.31\), \(A(18) = 31.29\)
      • The best whole number is 18, at $31.29 each.

Stretch

  1. The curve \(y = 0.5x^2 − 3x + 2\) and the line \(y = x − 3\) meet at points A and B.

    1234567−4−2246810xyAB
    1. (a) Find the coordinates of A and B.

      • Graph the curve and the line.
      • From the calculator: \(x = 1.55\) and \(x = 6.45\)
      • Use the line for \(y\): \(y = −1.45\) and \(y = 3.45\)
    2. (b) Find the length of AB.

      • \(AB = \sqrt{(6.45 − (1.55))^2 + (3.45 − (−1.45))^2}\)
      • \(AB = 6.93\)
  2. A ball is thrown upwards from the ground. Its height is \(h_1 = −4.9t^2 + 22t\) metres after \(t\) seconds. At the same moment a drone rises steadily from a height of 3 m, so its height is \(h_2 = 3 + 4t\) metres.

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

      • Graph \(y = −4.9x^2 + 22x\) and \(y = 3 + 4x\)
      • From the calculator: intersections at \(t = 0.175\) and \(t = 3.50\)
    2. (b) For how long is the ball higher than the drone?

      • The ball is higher between the two times.
      • \(3.50 − 0.175 = 3.32\) s

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