(a) Copy and complete the following table for the... | WAEC SSCE - General Mathematics - 1992 (Essay)

Question 1 Report

(a) Copy and complete the following table for the relation \(y = \frac{5}{2} + x - 4x^{2}\)

x	-2.0	-1.5	-1.0	-0.5	0	0.5	1	1.5	2.0
y	-15.5			1	2.5

(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the graph of the relation for \(-2.0 \leq x \leq 2.0\).

(d) From your graph, obtain the roots of the equation \(8x^{2} - 2x - 5 = 0\)

Answer

(a) Completing the table for \(y=\dfrac{5}{2}+x-4x^{2}\).

Each missing value is found by substitution, e.g.

\[ x=-1.5:\; y=2.5+(-1.5)-4(-1.5)^2 = 2.5-1.5-9 = -8.0 \] \[ x=1:\; y=2.5+1-4(1)^2 = 2.5+1-4 = -0.5 \] \[ x=1.5:\; y=2.5+1.5-4(1.5)^2 = 2.5+1.5-9 = -5.0 \] \[ x=2.0:\; y=2.5+2-4(2)^2 = 2.5+2-16 = -11.5 \]

x	-2.0	-1.5	-1.0	-0.5	0	0.5	1	1.5	2.0
y	-15.5	-8.0	-2.5	1.0	2.5	2.0	-0.5	-5.0	-11.5

(b) Graph of \(y=\dfrac{5}{2}+x-4x^{2}\) for \(-2.0\le x\le 2.0\). Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, the nine points from the table are plotted and joined with a smooth curve (an inverted parabola).

(c) Maximum value of y. The curve turns at its highest point where \(x=-\dfrac{b}{2a}=-\dfrac{1}{2(-4)}=0.125\). Reading the top of the curve from the graph:

\[ y_{\max}=2.5+0.125-4(0.125)^2 = 2.5+0.125-0.0625 \approx 2.6 \]

So the maximum value of \(y\) is approximately 2.6.

(d) Roots of \(8x^{2}-2x-5=0\). Divide the equation through by \(-2\):

\[ -4x^{2}+x+\tfrac{5}{2}=0 \quad\Longrightarrow\quad \tfrac{5}{2}+x-4x^{2}=0 \]

This is exactly \(y=0\), so the required roots are the values of \(x\) where the graph cuts the x-axis. Reading these two intercepts from the curve:

\[ x \approx -0.7 \qquad\text{and}\qquad x \approx 0.9 \]

(Check by formula: \(x=\dfrac{1\pm\sqrt{1+40}}{8}=\dfrac{1\pm 6.40}{8}\), giving \(x=-0.68\) or \(x=0.93\), which agrees with the graph.)

Read lesson note on Quadratic Equations (WAEC)

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