Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | June | Q#10
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 x-coordinate -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 x-coordinate 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 y-axis by finding the value of when
.
We can find the coordinates of y-intercept from the given equation of the curve.
Hence, the curve crosses y-axis at point .
ü Find the point(s) where the graph crosses the x-axis by finding the value of when
. If there is repeated root the graph will touch the x-axis.
We can find the coordinates of x-intercepts from the given equation of the curve.
Now we have two options.
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Hence, the curve crosses x-axis 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 y-axis.
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 (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).
First, we need y-coordinate 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 x-coordinate 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.
Point-Slope 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 x-coordinate of point B.
Now we have two options.
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We can deduce that the given curve has gradient 20 at two point where and x=-5.
We already know that point A has x-coordinate -5, therefore, point B has x-coordinate .
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