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

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.

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.

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:

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;

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;