# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2008 | January | Q#10

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Question

The curve C has equation

y=(x + 3)(x −1)2
.

a.   Sketch C showing clearly the coordinates of the points where the curve meets the coordinate  axes.

b.   Show that the equation of C can be written in the form

y = x3 + x− 5x + k,

where k is a positive integer, and state the value of k.

There are two points on C where the gradient of the tangent to C is equal to 3.

c.   Find the x-coordinates of these two points.

Solution

a.

We are given following function to sketch. It can be expanded to write as;   ü Find the sign of the coefficient of . This gives the shape of the graph at the extremities.

ü Find the point where the graph crosses y-axis by finding the value of when 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.

ü Calculate the values of for some value of . The 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.

Sign of the coefficient of is positive, hence, cubical graph increase as we move from left to right  on x-axis.

First we find coordinates of y-intercept of the graph.

The point at which curve (or line) intercepts y-axis, the value of . So we can find the  value of coordinate by substituting in the equation of the curve (or line).

We substitute in equation of the function to find the y-crossing of function.    Therefore, coordinates of y-intercept of the given graph are (0,3).

Now we find the x-intercepts of the graph by substituting

The point at which curve (or line) intercepts x-axis, the value of . So we can find the  value of coordinate by substituting in the equation of the curve (or line).

We substitute in equation of the function to find the y-crossing of function.  Now we have two options;        Two values of x indicate that there are two intersection points.

Therefore, graph intersects x-axis at two points and Now we can sketch the required graph as shown below. b.

We are given following function to sketch. It can be expanded to write as;   c.

We are given equation of the curve which can be written as demonstrated in (b); 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:    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 are given that slope of the curve is 3.

Therefore;     Now we have two options.         Therefore, slope of the curve is 3 at two points where;  