1
Graphs — reciprocal & quadratic curves
7 marks
a(i)
Complete the table of values for y = 4/x − 2/x² + 4, for −3 ≤ x ≤ 3. Values are given to 1 decimal place where appropriate.
[1]
| x | −3 | −2 | −1 | −0.5 | 0.25 | 0.5 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|---|---|
| y | ? | 1.5 | −2 | −12 | −12 | 4 | 6 | 5.5 | 5.1 |
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Concept: Substitution
Substitute x = −3 into y = 4/x − 2/x² + 4:
y = 4/(−3) − 2/(−3)² + 4 = −1.333 − 0.222 + 4
y = 2.4
a(ii)
On the grid, draw the graph of y = 4/x − 2/x² + 4 for −3 ≤ x ≤ 3.
[3]
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Concept: Plotting a curve
Plot each (x, y) pair from the completed table and join them with a smooth curve — the curve has two separate branches either side of the vertical asymptote at x = 0, since y is undefined when x = 0.
Curve plotted through all 9 points (see graph).
a(iii)
Write down the coordinates of the turning point on the graph of y = 4/x − 2/x² + 4.
[1]
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Concept: Reading a graph
Read off the highest point of the right-hand branch of the curve drawn in (a)(ii).
Turning point = (1, 6)
b
On the given axes, sketch the curve with equation y = −(x − 2)(x + 4). Indicate clearly the coordinates of any points where the curve crosses the coordinate axes.
[2]
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Concept: Sketching a quadratic from factorised form
x-intercepts: 0 = −(x−2)(x+4) → x = 2 or x = −4
y-intercept: substitute x = 0 → y = −(0−2)(0+4) = 8
x-coordinate of turning point = (x₁ + x₂)/2 = (2 + (−4))/2 = −1
Substitute x = −1: y = −[(−1)−2][(−1)+4] = 9
Downward parabola through (−4, 0), (2, 0) and (0, 8), with turning point (−1, 9).
