Figure 1 shows a sketch of the curve with equation
$y = \frac{3}{x}, \quad x \neq 0.$
(a) On a separate diagram, sketch the curve with equation
$y = \frac{3}{x+2}, \quad x \neq -2,$
showing the coordinates of any point at which the curve crosses a coordinate axis - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 1
Question 7
Figure 1 shows a sketch of the curve with equation
$y = \frac{3}{x}, \quad x \neq 0.$
(a) On a separate diagram, sketch the curve with equation
$y = \frac{3}{x... show full transcript
Worked Solution & Example Answer:Figure 1 shows a sketch of the curve with equation
$y = \frac{3}{x}, \quad x \neq 0.$
(a) On a separate diagram, sketch the curve with equation
$y = \frac{3}{x+2}, \quad x \neq -2,$
showing the coordinates of any point at which the curve crosses a coordinate axis - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 1
Step 1
Sketch the curve with equation $y = \frac{3}{x + 2}, \quad x \neq -2$
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Answer
Identify the Characteristics of the Curve:
The given function is similar to the original function y=x3, but is translated horizontally.
Finding the Asymptotes:
The vertical asymptote occurs where the denominator is zero. Therefore, for the equation y=x+23, the vertical asymptote is at
[ x + 2 = 0 \implies x = -2. ]
The horizontal asymptote is y=0, as the function approaches zero as x approaches infinity or negative infinity.
Intercepts:
To find the x-intercept, set y=0:
[ 0 = \frac{3}{x + 2} ]
which does not yield any valid x. Therefore, the curve does not intersect the x-axis.
To find the y-intercept, set x=0:
[ y = \frac{3}{0 + 2} = \frac{3}{2}. ]
Thus, the curve crosses the y-axis at the point ((0, \frac{3}{2})).
Sketch:
On a new diagram, sketch the curve, showing the asymptotes and the point of intersection at the y-axis.
Step 2
Write down the equations of the asymptotes of the curve in part (a)
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