A-Level Edexcel Maths Pure Inequalities: The curve C has equation $y = \f

The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 1

Question 8

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The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$. (a) On the axes below, sketch the graphs of C and l, indicating clearly the c... show full transcript

Worked Solution & Example Answer:The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 1

Step 1

a) On the axes below, sketch the graphs of C and l, indicating clearly the coordinates of any intersections with the axes.

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Answer

To sketch the curves, we need to find the intersections with the axes for both equations:

For curve C:

Setting $y = 0$ :

$\frac{3}{x} = 0$ → No intersection as $y$ never reaches zero.
Setting $x = 0$ :

This is undefined for $y = \frac{3}{x}$ .

For line l:

Setting $y = 0$ :

$2x + 5 = 0$ → $x = -\frac{5}{2}$ . This gives the point $(-\frac{5}{2}, 0)$ .
Setting $x = 0$ :

$y = 2(0) + 5 = 5$ . This gives the point $(0, 5)$ .

Sketch:

The graph of $C$ will have branches in the first and third quadrants and will asymptotically approach the axes.
The line $l$ will intersect the x-axis at $(-2.5, 0)$ and the y-axis at $(0, 5)$ .

Step 2

b) Find the coordinates of the points of intersection of C and l.

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Answer

To find the points of intersection:

Set the two equations equal:

$\frac{3}{x} = 2x + 5$ .
Multiply both sides by $x$ (condoning $x \neq 0$ ):

$3 = 2x^2 + 5x$ .

Rearranging gives us:

$2x^2 + 5x - 3 = 0$ .
Solve this quadratic using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{(5)^2 - 4(2)(-3)}}{2(2)}$

$= \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}$ .

This gives:

$x = \frac{2}{4} = \frac{1}{2}$ and $x = \frac{-12}{4} = -3$ .
Substitute these $x$ values back into either original equation to find corresponding $y$ values:

For $x = \frac{1}{2}$ :

$y = 2\left(\frac{1}{2}\right) + 5 = 1 + 5 = 6$ .

For $x = -3$ :

$y = 2(-3) + 5 = -6 + 5 = -1$ .
Intersection Points:

The coordinates of the points of intersection are:
- $igg(\frac{1}{2}, 6\bigg)$
- $(-3, -1)$ .

The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 1

Worked Solution & Example Answer:The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 1

a) On the axes below, sketch the graphs of C and l, indicating clearly the coordinates of any intersections with the axes.

b) Find the coordinates of the points of intersection of C and l.

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