WAEC 2013 · Paper 2 · Q7

  1. (a)

    Copy and complete the table of values for the relation y=3x2−5x−7y = 3x^2 - 5x - 7.

    xx −3-3 −2-2 −1-1 00 11 22 33 44
    yy 3535 −7-7 −9-9 55
    Model answer
    xx −3-3 −2-2 −1-1 00 11 22 33 44
    yy 3535 1515 11 −7-7 −9-9 −5-5 55 2121

    For example x=−2x = -2: 12+10−7=1512 + 10 - 7 = 15, and x=4x = 4: 48−20−7=2148 - 20 - 7 = 21.

  2. (b)

    Using scales of 2 cm to 1 unit on the xx-axis and 2 cm to 5 units on the yy-axis, draw the graph of y=3x2−5x−7y = 3x^2 - 5x - 7 for −3≤x≤4-3 \le x \le 4.

    Model answer
    −3−2−11234−10−55101520253035xy−0.92.6min ≈ −9.1(2, −5)y = 3x2 − 5x − 7tangent

    Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to 1 unit across, 2 cm to 5 units up.

    For (c): (i) roots x≈−0.9x \approx −0.9 and 2.62.6; (ii) the minimum is about −9.1−9.1 (at x≈0.8x \approx 0.8); (iii) draw the tangent at (2,−5)(2, -5) and measure its slope using two points on it far apart: the gradient is about 7.

  3. (c)

    From your graph: (i) find the roots of the equation 3x2−5x−7=03x^2 - 5x - 7 = 0; (ii) estimate the minimum value of yy; (iii) calculate the gradient of the curve at the point x=2x = 2.

    Separate values with commas, e.g. 3, −2

Try it on a graph

The curve and its tangent at x = 2 (gradient 7).

Worked solution (try it first)

(a)

  1. Substitute each xx into y=3x2−5x−7y = 3x^2 - 5x - 7.
  2. For example, x=−2x = -2 gives 12+10−7=1512 + 10 - 7 = 15 and x=4x = 4 gives 48−20−7=2148 - 20 - 7 = 21.
  3. xx −3-3 −2-2 −1-1 00 11 22 33 44
    yy 3535 1515 11 −7-7 −9-9 −5-5 55 2121

(b)

  1. With 2 cm to 1 unit on the xx-axis and 2 cm to 5 units on the yy-axis, plot the eight points and join them with a smooth U-shaped curve.

(c)(i)

  1. The roots are where the curve crosses the xx-axis: x≈−0.9x \approx -0.9 and x≈2.6x \approx 2.6.

(ii)

  1. The lowest point of the curve is a little right of x=1x = 1: the minimum value is y≈−9.1y \approx -9.1.

(iii)

  1. Draw the tangent at (2,−5)(2, -5).
  2. It passes through about (1,−12)(1, -12) and (3,2)(3, 2), so the rise is 2−(−12)=142 - (-12) = 14 and the run is 3−1=23 - 1 = 2.
  3. Its gradient is 14÷2=714 \div 2 = 7.
  4. (Check: dydx=6x−5=7\frac{dy}{dx} = 6x - 5 = 7 at x=2x = 2.)

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