WMA11 Jan 2019

11. WMA11/01 Edexcel IAL P1 January 2019, Q11 (Graphs and Transformation: Sketching Graphs)

(a) On Diagram 1 sketch the graphs of

(i) y = x(3 − x)

(ii) y = x(x − 2)(5 − x)

showing​​ clearly the coordinates of the points where the curves cross the coordinate axes.

(4)

(b) Show that the x coordinates of the points of intersection of

y = x(3 - x) and y = x(x - 2)(5 - x)

​​ are given by the solutions to the equation x(x2 − 8x + 13) = 0

(3)

The point P lies on both curves. Given that P lies in the first quadrant,

(c) find, using algebra and showing your working, the exact coordinates of P.

(5)

SOLUTION​​ 

a- i-​​ y = x(3 − x)

For x-intercepts or solutions of y, put y=0.​​ 

x3-x=0

x=0& x=3

Also, this is the equation of a quadratic curve as on​​ multiplying ​​ x with (3-x), the maximum power of x would be 2. And coefficient of x2​​ would be -1. Hence, the graph would be cup-down.​​ 

a- ii-​​ y = x(x − 2)(5 − x)

For x-intercepts or solutions of y, put y=0.​​ 

y = x(x - 2)(5 - x)

0= x(x - 2)(5 - x)

x=0,  x=2,  x=5

Also, this is the equation of a cubic curve as​​ on multiplying ​​ x with the brackets, the maximum power of x would be 3. And coefficient of x3​​ would be -1. Hence, the graph would have first minimum point and then maximum point. ​​ 

b-​​ 

xx-25-x=x3-x

x5x-x2-10+2x=3x-x2

5x2-x3-10x+2x2=3x-x2

-x3+8x2-13x=0

x3-8x2+13x=0

xx2-8x  13=0

c-​​ Since P lies on both curves, it must be a point of intersection of the curves. And, it lies in the first quadrant which means the value of solutions must be not less than or equals to zero.​​ 

xx2-8x+13=0

x=0                     x2-8x+13=0

Now, solving for​​ 

x2-8x+13=0

x2-8x=-13

x-42-16=-13

x-42=3

x-4= ±3

x= 4-3

So, the value of x will be​​ 4-3​​ since it is greater than 0.​​ 

Whereas the ​​ value of y would be​​ 

y=4-33-4-3

y=4-3-1+3

y=-4+43+3-3

y=-7+53

Hence, the coordinates of P are​​ 

(4-3,-7+53)