A-Level Edexcel Maths Pure Graphs of Functions: 10. (a) On the axes below, sketc

10. (a) On the axes below, sketch the graphs of (i) $y = x(x + 2)(3 - x)$ (ii) $y = -\frac{2}{x}$ showing clearly the coordinates of all the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2011 - Paper 2

Question 11

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10. (a) On the axes below, sketch the graphs of (i) $y = x(x + 2)(3 - x)$ (ii) $y = -\frac{2}{x}$ showing clearly the coordinates of all the points where the cu... show full transcript

Worked Solution & Example Answer:10. (a) On the axes below, sketch the graphs of (i) $y = x(x + 2)(3 - x)$ (ii) $y = -\frac{2}{x}$ showing clearly the coordinates of all the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2011 - Paper 2

Step 1

(i) $y = x(x + 2)(3 - x)$

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Answer

To find the points where this curve crosses the coordinate axes, we start by determining the x-intercepts by setting $y = 0$ :

$x(x + 2)(3 - x) = 0$

This gives us:

$x = 0$ ,
$x + 2 = 0 \Rightarrow x = -2$ ,
$3 - x = 0 \Rightarrow x = 3$ .

Thus, the points where it crosses the x-axis are (0, 0), (-2, 0), and (3, 0).

Next, to find the y-intercept, we evaluate $y$ when $x = 0$ :

$y = 0(0 + 2)(3 - 0) = 0$

So the curve also crosses the y-axis at (0, 0). The curve is a cubic shape, and we ensure it passes through these points correctly.

Step 2

(ii) $y = -\frac{2}{x}$

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Next, to find where this curve crosses the axes:

For the x-intercept, setting $y = 0$ gives no solution, as $-\frac{2}{x} = 0$ has no roots. This means the curve does not cross the x-axis.

For the y-intercept, we evaluate at $x = 1$ (or any other non-zero):

$y = -\frac{2}{1} = -2$

So the curve crosses the y-axis at (0, -2). The graph has two branches, located in the second and fourth quadrants, asymptotic to the axes and showing a negative trend.

Step 3

Using your sketch state, giving a reason, the number of real solutions to the equation

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For the equation $x(x + 2)(3 - x) + \frac{2}{x} = 0$ , we have to consider the graphs of both functions: $y = x(x + 2)(3 - x)$ and $y = -\frac{2}{x}$ .

Using the earlier sketches, we note that:

The cubic function has three x-intercepts: (0, 0), (-2, 0), and (3, 0).
The hyperbola intersects the y-axis at (0, -2) and does not touch the x-axis.

By examining where these graphs intersect, we find there are 2 points of intersection. Thus, we conclude that there are two real solutions to the equation.

10. (a) On the axes below, sketch the graphs of (i) $y = x(x + 2)(3 - x)$ (ii) $y = -\frac{2}{x}$ showing clearly the coordinates of all the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2011 - Paper 2

Worked Solution & Example Answer:10. (a) On the axes below, sketch the graphs of (i) $y = x(x + 2)(3 - x)$ (ii) $y = -\frac{2}{x}$ showing clearly the coordinates of all the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2011 - Paper 2

(i) $y = x(x + 2)(3 - x)$

(ii) $y = -\frac{2}{x}$

Using your sketch state, giving a reason, the number of real solutions to the equation

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