Figure 1 shows a sketch of the curve with equation $y = \frac{2}{x}$, $x \neq 0$. The curve C has equation $y = \frac{2}{x} - 5$, $x \neq 0$, and the line l has equation $y = 4x + 2$. (a) Sketch and clearly label the graphs of C and l on a single diagram. On your diagram, show clearly the coordinates of the points where C and l cross the coordinate axes. (b) Write down the equations of the asymptotes of the curve C. (c) Find the coordinates of the points of intersection of $y = \frac{2}{x} - 5$ and $y = 4x + 2$.

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Figure 1 shows a sketch of the curve with equation $y = \frac{2}{x}$, $x \neq 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 3

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Figure 1 shows a sketch of the curve with equation $y = \frac{2}{x}$, $x \neq 0$. The curve C has equation $y = \frac{2}{x} - 5$, $x \neq 0$, and the line l has equ... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of the curve with equation $y = \frac{2}{x}$, $x \neq 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 3

Step 1

Sketch and clearly label the graphs of C and l on a single diagram.

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Answer

To illustrate the graphs of both curves, we start by plotting the graph of the curve $C$ , which is defined by the function $y = \frac{2}{x} - 5$ . This function has a vertical asymptote at $x = 0$ and a horizontal asymptote at $y = -5$ . The graph of the line $l$ defined by $y = 4x + 2$ will be a straight line that crosses the y-axis at $(0, 2)$ and has a slope of 4.

Coordinates of Intersections:
- For curve $C$ $C$ :
  - $x$ -intercept: Set $y = 0$ : $0 = \frac{2}{x} - 5 \implies \frac{2}{x} = 5 \implies x = \frac{2}{5}$
    Thus, the x-intercept is at $\, (\frac{2}{5}, 0)$ .
- For line $l$ $l$ :
  - $x$ -intercept: Set $y = 0$ : $0 = 4x + 2 \implies 4x = -2 \implies x = -\frac{1}{2}$ Thus, the x-intercept is at $( -\frac{1}{2}, 0)$ .
The y-intercept of $C$ $C$ can be found by substituting $x$ $x$ :
- $y = \frac{2}{1} - 5 = -3$ Thus, the y-intercept is at $(0, -3)$ .

Step 2

Write down the equations of the asymptotes of the curve C.

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Answer

The equations of the asymptotes for curve C are:

Vertical Asymptote: $x = 0$
Horizontal Asymptote: $y = -5$

Step 3

Find the coordinates of the points of intersection of $y = \frac{2}{x} - 5$ and $y = 4x + 2$.

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Answer

To find the points of intersection, we set the functions equal to each other: $\frac{2}{x} - 5 = 4x + 2$

Rearranging gives: $\frac{2}{x} = 4x + 7$
Multiplying both sides by $x$ (where $x \neq 0$ ) results in: $2 = 4x^2 + 7x$
Rearranging results in: $4x^2 + 7x - 2 = 0$
Applying the quadratic formula, we have: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}$
This simplifies to: $x = \frac{-7 \pm \sqrt{49 + 32}}{8} = \frac{-7 \pm \sqrt{81}}{8} = \frac{-7 \pm 9}{8}$

Thus, we find two potential x values:

$x_1 = \frac{2}{8} = \frac{1}{4}$
$x_2 = \frac{-16}{8} = -2$

Now substituting back to find corresponding $y$ values:

For $x = \frac{1}{4}$ : $y = 4(\frac{1}{4}) + 2 = 1 + 2 = 3$
For $x = -2$ : $y = 4(-2) + 2 = -8 + 2 = -6$

Thus, the points of intersection are:

$\left( \frac{1}{4}, 3 \right)$ and $(-2, -6)$ .

Figure 1 shows a sketch of the curve with equation $y = \frac{2}{x}$, $x \neq 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 3

Worked Solution & Example Answer:Figure 1 shows a sketch of the curve with equation $y = \frac{2}{x}$, $x \neq 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 3

Sketch and clearly label the graphs of C and l on a single diagram.

Write down the equations of the asymptotes of the curve C.

Find the coordinates of the points of intersection of $y = \frac{2}{x} - 5$ and $y = 4x + 2$.

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