WMA11 January 2020

10. WMA11/01​​ Edexcel IAL P1​​ January 2020, Q10 (Graphs &​​ Transformations)

The curve C1​​ has equation y = f(x), where​​ 

f(x) = 4x – 3x – 52

(a) Sketch C1​​ showing the coordinates of any point where the curve touches or crosses the coordinate axes.​​ 

(3)

(b) Hence or otherwise​​ 

• find the values of x for which​​ f14x=0 ​​ 

• find the value of the constant p such that the curve with equation y = f(x) + p passes through the origin.​​ 

(2)

A second curve C2​​ has equation y = g(x), where g(x) = f(x + 1)​​ 

(c)

• Find, in simplest form, g(x). You may leave your answer in a factorised form.​​ 

• Hence, or otherwise, find the y intercept of curve C2​​ 

(3)

SOLUTION

a-​​ ​​ 

For y-intercept of cubic graph, put x=0 in the equation of a curve.​​ 

y = 4x – 3x – 52

y=0-30-52

y= -3 25

y=-75

So, the y-intercept is (0, -75).

Whereas the solution of the cubic equation can be found by following method.​​ 

4x-3x-52=0

4x-3x-5x-5=0

4x-3=0            x=5 repeated

x=0.75

Hence, the sketch​​ of cubic curve looks like the following.​​ 

b-i-​​ 

f(x) = 4x – 3x – 52

f14x=414x-314x-52

f14x=x-314x-52

Finding value of x when​​ 

f14x=0

x-314x-52=0

x-3=0     or     14x-52=0

x=3     or     14x-52=0

x=3     or     14x-5=0

14x=5

                         x=20

b- ii-​​ 

y=f x+p

y=4x-3x-52+p

As the curve​​ y = f(x) + p passes through origin.​​ 

0=-3-52+p

0=-75+p

p=75

c- i-

gx=fx+1

gx=4x+1-3x+1-52

gx=4x+1x-42

c- ii-​​ 
For intercept, put x=0.

gx=4x+1x-42

gx=4×0+10-42

y=1-42

y=1 x 16

y=16