Question

The curve  has equation

a.   Find .

b.   Sketch , showing the coordinates of the points where C1 meets the x-axis.

c.   Find the gradient of  at each point where C1 meets the x-axis.

The curve  has equation

where k is a constant and .

d.   Sketch , showing the coordinates of the points where  meets the x and y axes.

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  with respect to  is:

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

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 . 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.

c.
 

We are required o find the gradient of the curve at each point it meets the x-axis.

The curve meets x-axis at two points as demonstrated in (b);  and .

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)  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;

Therefore, we need .

From (a) we already have found that;

For gradient of the curve at points (0,0), substitute  in derivative of the equation of the curve. 

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)  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;

Therefore, we need .

From (a) we already have found that;

For gradient of the curve at points (0,0), substitute  in derivative of the equation of the curve. 

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  transforms the graph of  into the graph of . 

Transformation of the function  into  results from translation through vector  .

Translation through vector  represents the move,  units in the positive x-direction and  units in the positive y-direction.

	Original	Transformed	Translation Vector	Movement
Function				units in positive x-direction  units in positive y-direction
Coordinates

However, for the given case we consider following.

Translation through vector represents the move,  units in the positive x-direction and  units in the y-direction.

Translation through vector  transforms the function  into .

Transformation of the function  into  results from translation through vector . 

Translation through vector  transforms the function  into  which means shift  towards right along x-axis.

	Original	Transformed	Translation Vector	Movement
Function				units in positive x-direction
Coordinates

It is evident that we are required to transform the function  into , therefore it is  case of translation of  along negative x-axis by k units. 

It is also evident from the above table that only x-coordinates of the graph change whereas y- coordinates of the graph will remain unchanged.

Hence, the new function has all the y-coordinates same as that of original given function whereas  all the x-coordinates are shifted towards positive x-axis of original given function.

Since k>2, we shift graph of original curve as sketched in (b), towards positive x-axis at least 3  units.

However, we need to mark the y-intercept of the curve as well.

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 have equation of the curve  as; 

We substitute x=0 to find the coordinates of y-intercept.

Hence, coordinates of the y-intercept of the curve  are .