Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2017 | June | Q#10



Figure shows a sketch of part of the curve y = f(x), , where

f(x) = (2x – 5)2(x + 3)

a.   Given that

                    i.       the curve with equation y = f(x) – k, , passes through the origin, find the value of the  constant k,

                  ii.       the curve with equation y = f(x + c), , has a minimum point at the origin, find the value  of the constant c.

b.   Show that

Points A and B are distinct points that lie on the curve y = f(x).

The gradient of the curve at A is equal to the gradient of the curve at B.

Given that point A has x coordinate is 3.

c.   find the x coordinate of point B.




We are given that equation of the curve is;

We are also given that  passes through origin. Therefore, O(0,0) lies on the curve with  equation;

If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the  curve  (or the line).

Therefore, we can substitute the coordinates of origin in given equation of the curve and it must  satisfy the equation.


We are given that equation of the curve is;

We are also given that  has a minimum point at the origin.

First we find the coordinates of the minimum point for the given curve with equation;

We can see from the diagram that minimum point is also the x-intercept of the curve.

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).

Therefore, we substitute  in equation of the curve.

Now we have two options.

It is evident from the diagram that  does not represent the minimum point of the curve  because minimum point of the curve is on positive side of the x-axis. Hence, minimum point has x- coordinate .

The given equation  represents translation of graph along x-axis.

Translation through vector  represents the move,  units in the negative 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 left along x-axis. 







 units in
negative x-direction


If we need to translate the graph such that minimum point shifts to origin, it must move to the left  side (towards negative x-axis) equal to x-coordinate of minimum point ie . Hence;


We are required to find the expression for gradient of the curve.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:


Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:


We are given that the gradient of the curve at A is equal to the gradient of the curve at B.

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 also given that x-coordinates of the point A is 3. Therefore, we can find gradient of the curve  at point A.

From (b) we have found that;



Therefore, as per given condition;

We can also write as;

We can solve this equation to find the x-coordinates of all such points where gradient of the curve is  25.

Now we have two options.

Therefore, at two points (one with x-coordinate 3 and other with ) the gradient of the curve is 25  and we are given that these two points are A and B.

We are given that x-coordinate of point A is 3, therefore,  must be x-coordinate of the point B.