On separate axes sketch the graphs of
(i) $y = -3x + c$, where $c$ is a positive constant,
(ii) $y = \frac{1}{x} + 5$
On each sketch show the coordinates of any point at which the graph crosses the $y$-axis and the equation of any horizontal asymptote - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 1
Question 2
On separate axes sketch the graphs of
(i) $y = -3x + c$, where $c$ is a positive constant,
(ii) $y = \frac{1}{x} + 5$
On each sketch show the coordinates of any ... show full transcript
Worked Solution & Example Answer:On separate axes sketch the graphs of
(i) $y = -3x + c$, where $c$ is a positive constant,
(ii) $y = \frac{1}{x} + 5$
On each sketch show the coordinates of any point at which the graph crosses the $y$-axis and the equation of any horizontal asymptote - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 1
Step 1
Sketch the graph of $y = -3x + c$
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Answer
The graph of the equation is a straight line with a negative gradient.
It crosses the y-axis at the point (0,c), where c is a positive constant.
Since the slope is negative, the line will decrease as x increases.
Step 2
Sketch the graph of $y = \frac{1}{x} + 5$
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Answer
The graph has two branches with a horizontal asymptote at y=5. This is because as x approaches infinity, x1 approaches 0, making y approach 5.
The graph crosses the y-axis at (0,5) since plugging x=0 into y=x1+5 is undefined, indicating an asymptote at x=0.
The branches will approach the horizontal asymptote near y=5.
Step 3
show that $(5 - c^2) > 12$
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Answer
Starting from the equation:
x1+5=−3x+c
Rearranging gives:
5−c=−3x−x1
Squaring both sides:
(5−c)2=(−3x−x1)2
Solving gives:
b2−4ac⇒b2−4ac=(5−c)2−4(1)(12)>0
Step 4
Hence find the range of possible values for $c$
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Answer
From the inequality (5−c2)>12, we can rearrange it to yield: 5−c2>12−c2>7c2<−7c<−7 or c>7.
Therefore, the possible values for c are:
c<−7 or c>7, excluding values where c is negative, hence the valid range is c>7.