Tomas grows vegetables on a garden allotment and writes a page for the allotment club newsletter. For the page he draws the graph of \(y = x^2 - 3x - 2\) fo...

Assessment: Mathematics Specification B 4MB1 | Paper 2 Mock 01 | Written Paper 2 Subject: Mathematics Specification B - 4MB1

Question 1 Report

Tomas grows vegetables on a garden allotment and writes a page for the allotment club newsletter. For the page he draws the graph of \(y = x^2 - 3x - 2\) for values of \(x\) from \(-2\) to 5. The table shows some of the values he has worked out. Two of the values are missing. An empty grid is given below the table.

x-2-1012345
y8-2-4-428
-2-112345-5-4-3-2-1123456789xy© EAGLE BEACON GLOBAL
  1. Complete the table of values. (2)
  2. On the grid, draw the graph of \(y = x^2 - 3x - 2\) for \(-2 \leq x \leq 5\). (2)
  3. Use your graph to find estimates for the two solutions of \(x^2 - 3x - 2 = 0\). (2)
  4. Write down the coordinates of the minimum point of the curve. (1)

Answer Details

Drawing a quadratic graph accurately depends on getting every table value right, since one wrong point distorts the whole curve. The graph then gives estimates for the roots and shows the minimum point.

(a) Complete the table. [2]
Substitute into \(y = x^2 - 3x - 2\), taking care with the signs when \(x\) is negative:

\[x = -1: \quad (-1)^2 - 3(-1) - 2 = 1 + 3 - 2 = 2\] \[x = 3: \quad 3^2 - 3(3) - 2 = 9 - 9 - 2 = -2\]
\(x\)-2-1012345
\(y\)82-2-4-4-228

Note \(-3 \times (-1) = +3\); treating it as \(-3\) gives \(y = -4\) and spoils the shape of the curve.

(b) Draw the graph. [2]
Plot the eight points and join them with a single smooth curve, not a series of straight segments:

-224-4-2268xy© EAGLE BEACON GLOBAL

(c) Estimates for the solutions of \(x^2 - 3x - 2 = 0\). [2]
The solutions are the \(x\) values where the curve crosses the horizontal axis, marked above. Reading off:

\[x \approx -0.6 \qquad\text{and}\qquad x \approx 3.6\]

The quadratic formula confirms these: \(x = \dfrac{3 \pm \sqrt{9 + 8}}{2} = \dfrac{3 \pm \sqrt{17}}{2}\), giving \(-0.561\ldots\) and \(3.561\ldots\)

(d) Minimum point. [1]
The table shows equal values of \(-4\) at \(x = 1\) and \(x = 2\), so by symmetry the lowest point is halfway between them, at \(x = 1.5\):

\[y = (1.5)^2 - 3(1.5) - 2 = 2.25 - 4.5 - 2 = -4.25\]

The minimum point is \((1.5,\ -4.25)\).

Examination point: when two table values are equal, the turning point lies exactly halfway between those two \(x\) values. That is faster and more accurate than trying to read the lowest point of a hand-drawn curve, and it also gives the line of symmetry, \(x = 1.5\).

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