Using a scale of 2cm to 1 unit on the x- axis and 1cm to 1 unit on the y- axis, draw on the same axes the graphs of \(y = 3 + 2x - x^{2}; y = 2x - 3\) for \...
Question 1 Report
Using a scale of 2cm to 1 unit on the x- axis and 1cm to 1 unit on the y- axis, draw on the same axes the graphs of \(y = 3 + 2x - x^{2}; y = 2x - 3\) for \(-3 \leq x \leq 4\). Using your graph:
(i) solve the equation \(6 - x^{2} = 0\);
(ii) find the maximum value of \(3 + 2x - x^{2}\);
(iii) find the range of x for which \(3 + 2x - x^{2} \leq 1\), expressing all your answers correct to one decimal place.
Answer Details
Table of values
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| \(y=3+2x-x^2\) | -12 | -5 | 0 | 3 | 4 | 3 | 0 | -5 |
| \(y=2x-3\) | -9 | -7 | -5 | -3 | -1 | 1 | 3 | 5 |
Using the stated scales, the required graph is:
(i) At the points of intersection,
\[3+2x-x^2=2x-3\]
\[6-x^2=0.\]
Reading the two intersection abscissae from the graph gives
\[x\approx -2.4\quad\text{or}\quad x\approx 2.4.\]
(ii) The highest point of the parabola is \((1,4)\). Hence, the maximum value of \(3+2x-x^2\) is
\[\boxed{4.0}.\]
(iii) The curve \(y=3+2x-x^2\) meets the line \(y=1\) at approximately \(x=-0.7\) and \(x=2.7\). The part of the parabola on or below \(y=1\) is therefore
\[\boxed{-3.0\leq x\leq -0.7\quad\text{or}\quad 2.7\leq x\leq4.0}.\]
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