Figure 1 shows a sketch of the curve with equation
$$y = \frac{2}{x}, \quad x \neq 0$$
The curve C has equation
$$y = \frac{2}{x} - 5, \quad x \neq 0$$
and the line l has equation
$$y = 4x + 2.$$
(a) Sketch and clearly label the graphs of C and l on a single diagram - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 3
Question 7
Figure 1 shows a sketch of the curve with equation
$$y = \frac{2}{x}, \quad x \neq 0$$
The curve C has equation
$$y = \frac{2}{x} - 5, \quad x \neq 0$$
and the lin... show full transcript
Worked Solution & Example Answer:Figure 1 shows a sketch of the curve with equation
$$y = \frac{2}{x}, \quad x \neq 0$$
The curve C has equation
$$y = \frac{2}{x} - 5, \quad x \neq 0$$
and the line l has equation
$$y = 4x + 2.$$
(a) Sketch and clearly label the graphs of C and l on a single diagram - 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
Sketch the curve C: Start by plotting the hyperbola defined by the equation y=x2. The curve will approach the x-axis (y = 0) and y-axis (x = 0) as asymptotes.
Shift the hyperbola: Since C’s equation is y=x2−5, shift the entire hyperbola downward by 5 units. The new asymptote is now at y = -5.
Sketch the line l: Draw the line given by the equation y=4x+2. This line has a y-intercept at (0, 2) and a slope of 4.
Label axes crossings: Identify points where C crosses the x-axis (i.e., set y = 0) and where l crosses the axes. Ensure these points are marked clearly on the diagram.
Step 2
Write down the equations of the asymptotes of the curve C.
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Answer
The asymptotes for the curve C given by (y = \frac{2}{x} - 5) are:
The vertical asymptote at (x = 0)
The horizontal asymptote at (y = -5)
Step 3
Find the coordinates of the points of intersection of y = 2/x - 5 and y = 4x + 2.
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Answer
To find the points of intersection, set the equations equal: x2−5=4x+2
Multiply through by x to eliminate the fraction: 2−5x=4x2+2x
Rearranging gives: 4x2+7x−2=0
Use the quadratic formula to solve for x: x=2a−b±b2−4ac=8−7±49+32=8−7±9
This gives two potential solutions:
(x = \frac{1}{4} \approx 0.25)
(x = -2)
Now, substitute these x-values back into either original equations to obtain corresponding y-values:
For (x = \frac{1}{4}): y=4(41)+2=3
For (x = -2): y=4(−2)+2=−6
The points of intersection are approximately (\left(\frac{1}{4}, 3\right)) and ((-2, -6)).