(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \(y - \frac{3x}{4} = 3\) and \(y + 2x = 6\). (ii) From you...
Question 1 Report
(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \(y - \frac{3x}{4} = 3\) and \(y + 2x = 6\).
(ii) From your graph, find the coordinates of the point of intersection of the two graphs.
(iii) Show, on the graph sheet, the region satisfied by the inequality \(y - \frac{3}{4}x \geq 3\).
(b) Given that \(x^{2} + bx + 18\) is factorized as \((x + 2)(x + c)\). Find the values of c and b.
Answer Details
(a)(i) Rearrange the equations:
\[y-\frac{3x}{4}=3\quad\Rightarrow\quad y=\frac34x+3\]
\[y+2x=6\quad\Rightarrow\quad y=6-2x\]
| Line | Points used for plotting |
|---|---|
| \(y=\frac34x+3\) | \((-4,0),\ (0,3),\ (4,6)\) |
| \(y=6-2x\) | \((0,6),\ (1,4),\ (3,0)\) |
Using the scale of 2 cm to 1 unit on both axes, the required graph is:
The boundary \(y-\frac34x=3\) is drawn as a solid line since the inequality includes equality. Region \(R\), the side above this line, is shaded.
(ii) The two lines intersect at approximately
\[\boxed{(1.1,\ 3.8)}\]
This agrees with the exact solution:
\[\frac34x+3=6-2x\]
\[\frac{11}{4}x=3\quad\Rightarrow\quad x=\frac{12}{11}\]
\[y=6-2\left(\frac{12}{11}\right)=\frac{42}{11}\]
Thus the exact coordinates are \(\left(\frac{12}{11},\frac{42}{11}\right)\), which are approximately \((1.1,3.8)\).
(iii) \[y-\frac34x\geq3\quad\Rightarrow\quad y\geq\frac34x+3.\]
Hence, the required region is the half-plane on and above \(y=\frac34x+3\), labelled \(R\) on the graph.
(b)
\[(x+2)(x+c)=x^2+(c+2)x+2c.\]
Comparing with \(x^2+bx+18\):
\[2c=18\Rightarrow c=9,\]
\[b=c+2=9+2=11.\]
\[\boxed{c=9,\quad b=11}\]
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