Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2012  January  Q#8
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Question
The curve has equation
a. Find
b. Sketch
c. Find the gradient of
The curve
where k is a constant and
d. Sketch
Solution
a.
We are given;
We can rewrite it as;
We are required to find
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve
Therefore;
Rule for differentiation is of
Rule for differentiation is of
b.
We are required to sketch;
As demonstrated in (a), we can write it as;
It is evident that it is a cubic equation.
We can now sketch the curve as follows.
ü Find the sign of the coefficient of
It is evident that with positive coefficient of
ü Find the point where the graph crosses yaxis by finding the value of
We can find the coordinates of yintercept from the given equation of the curve.
Hence, the curve crosses yaxis at point
ü Find the point(s) where the graph crosses the xaxis by finding the value of
We can find the coordinates of xintercepts from the given equation of the curve.
Now we have two options.






Hence, the curve crosses xaxis at points
ü Calculate the values of
ü Complete the sketch of the graph by joining the sections.
ü Sketch should show the main features of the graph and also, where possible, values where the graph intersects coordinate axes.
c.
We are required o find the gradient of the curve at each point it meets the xaxis.
The curve meets xaxis at two points as demonstrated in (b);
First we find the gradient of the curve at point (0,0).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope)
Therefore, we need
From (a) we already have found that;
For gradient of the curve at points (0,0), substitute
Hence, gradient of the curve at points (0,0) is;
Next we find the gradient of the curve at point (2,0).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope)
Therefore, we need
From (a) we already have found that;
For gradient of the curve at points (0,0), substitute
Hence, gradient of the curve at points (2,0) is;
d.
We are required to sketch;
It is evident that we have a translated version of the equation of curve
Translation through vector
Transformation of the function
Translation through vector
Original 
Transformed 
Translation Vector 
Movement 

Function 




Coordinates 


However, for the given case we consider following.
Translation through vector
Translation through vector
Transformation of the function
Translation through vector
Original 
Transformed 
Translation Vector 
Movement 

Function 




Coordinates 


It is evident that we are required to transform the function
It is also evident from the above table that only xcoordinates of the graph change whereas y coordinates of the graph will remain unchanged.
Hence, the new function has all the ycoordinates same as that of original given function whereas all the xcoordinates are shifted towards positive xaxis of original given function.
Since k>2, we shift graph of original curve as sketched in (b), towards positive xaxis at least 3 units.
However, we need to mark the yintercept of the curve as well.
The point
We have equation of the curve
We substitute x=0 to find the coordinates of yintercept.
Hence, coordinates of the yintercept of the curve
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