A-Level Edexcel Maths Pure Equation of a Straight Line: 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 x-intercepts, set the equation equal to zero: $x(x + 2)(3 - x) = 0$ The solutions are:

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

Thus, the x-intercepts are at the points: $(0, 0)$ , $(-2, 0)$ , and $(3, 0)$ .

To find the y-intercept, set $x = 0$ : $y = 0(0 + 2)(3 - 0) = 0$ The y-intercept is at the point $(0, 0)$ .

The cubic function has the shape of a downwards-facing cubic curve, passing through these points.

Step 2

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

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To find the x-intercepts, set the equation equal to zero: $-\frac{2}{x} = 0$ This equation has no solutions since a fraction cannot equal zero. Hence, there are no x-intercepts.

To find the y-intercept, set $x = 1$ (a test point): $y = -\frac{2}{1} = -2$ Thus, the y-intercept is at the point $(1, -2)$ .

The hyperbola branches in quadrants II and IV without crossing the axes.

Step 3

Using your sketch state, giving a reason, the number of real solutions to the equation $x(x + 2)(3 - x) + \frac{2}{x} = 0$

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Examining the equation: $x(x + 2)(3 - x) + \frac{2}{x} = 0$ From the sketch, the curve of $y = x(x + 2)(3 - x)$ intersects the horizontal line of $y = -\frac{2}{x}$ at two points. This means there are '2' real solutions to the equation as indicated by the intersections, demonstrating two places where the graphs meet.

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 $x(x + 2)(3 - x) + \frac{2}{x} = 0$

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