Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2011  June  Q#10
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Question
The curve C has equation
y = (x +1)(x + 3)^{2}
a. Sketch C, showing the coordinates of the points at which C meets the axes.
b. Show that .
The point A, with xcoordinate 5, lies on C.
c. Find the equation of the tangent to C at A, giving your answer in the form y = mx + c, where m and c are constants.
Another point B also lies on C. The tangents to C at A and B are parallel.
d. Find the xcoordinate of B.
Solution
a.
We are required to sketch;
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 . This gives the shape of the graph at the extremities.
It is evident that with positive coefficient of will shape the curve at extremities like increasing from left to right.
ü Find the point where the graph crosses yaxis by finding the value of when .
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 when . If there is repeated root the graph will touch the xaxis.
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 and .
ü Calculate the values of for some value of . This is particularly useful in determining the quadrant in which the graph might turn close to the yaxis.
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.
b.
We are given;
We are required to find .
As demonstrated in (a), we can write it as;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Therefore;
Rule for differentiation is of is:
Rule for differentiation is of is:
Rule for differentiation is of is:
c.
We are required to find the equation of tangent to the curve C at point A(5,y).
To find the equation of the line either we need coordinates of the two points on the line (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
First, we need ycoordinate of the point A.
If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve (or the line).
We are given that point A lies on the curve C which has equation;
We substitute xcoordinate of point A in the equation of the curve C;
Hence, coordinates of point A are (5,16).
Next, we need to find slope of tangent at in order to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we can find slope of the curve C at point A(5,16). then we can find slope of the tangent to the curve at this point.
We need to find the gradient of the curve C at point A(5,16).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
We have already found in (b);
For gradient of the curve at point A(5,y), substitute in derivative of the equation of the curve.
Therefore;
With coordinates of a point on the tangent A(5,16) and its slope in hand, we can write equation of the tangent.
PointSlope form of the equation of the line is;
d.
We are given that another point B also lies on C and the tangents to C at A and B are parallel.
If two lines are parallel to each other, then their slopes and are equal;
Therefore;
We have found in (c) that;
Hence;
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore;
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
We have found in (b) that expression for slope of the given curve is;
Therefore;
We can solve this equation to find the xcoordinate of point B.
Now we have two options.









We can deduce that the given curve has gradient 20 at two point where and x=5.
We already know that point A has xcoordinate 5, therefore, point B has xcoordinate .
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