WMA11 October 2021

6. WMA11/01 Edexcel IAL P1 Oct 2021, Q6 (Graphs &​​ Transformation: Cubic Curves and Differentiation)

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

A curve C has equation y = f(x) where​​ 

f(x) = 2x + 1x – 32

(a) Sketch a graph of C.​​ 

Show on your graph the coordinates of the points where C cuts or meets the coordinate axes.

(3)

(b) Write f(x) in the form​​ ax3 + bx2 + cx + d, where a, b, c and d are constants to be found.​​ 

(3)

(c) Hence, find the equation of the tangent to C at the point where​​ x =13​​ 

(4)

SOLUTION​​ 

a- Let us first check for y-intercept, where​​ x=0.​​ 

 y=21-32=219

y=18

Now, let us check for x-intercept, where y=0

2x+1x-32=0

x+1=0       x-3=0    

x=-1  &   x=3 

Since on expanding the brackets of the given equation of a curve, the coefficient of x3​​ is 2>0, the shape of cubic graph would be​​ 

b-

fx=2x+2x2-6x+9

f(x)=2x3-12x2+18x+2x2-12x+18

fx=2x3 -10x2+6x+18 

Hence,​​ a=2, b=-10, c=6, and​​ d=18.​​ 

c-

The equation of a function is​​ 

fx=2x3 -10x2+6x+18

The equation of tangent of the curve is found by differentiating the equation of function.

f'x=6x2-20x+6

Now, substituting the value of x =1/3.​​ 

f'13=6132-2013+6

f'13=0

Hence, the gradient of tangent is mT=0, meaning thereby the tangent is horizontal.

When x=1/3, the value of y is​​ 

f13=213+113-32

y=243-932

y=51227

This means the tangent is horizontal and it passes through a point (1/3, 512/27). So on putting the values, we get​​ 

y=mx+c

51227=013+c

c=51227

Hence, the equation of tangent at x=1/3 is​​ 

y=mx+c

y=013+51227

y=51227