WMA11 Jan 2021

6. WMA11/01 Edexcel IAL P1 January 2021 IAL Q6 (Graphs and Transformations: Reciprocal Graphs &​​ Points of Intersection)

• Sketch the curve with equation​​ 

y=-kx       k>0        x≠0 

(2)

• On a separate diagram, sketch the curve with equation​​ 

y=-kx+k       k>0        x≠0 

stating the coordinates of the point of intersection with the x-axis and, in terms of k, the equation of the horizontal asymptote.​​ 

(3)

• Find the range of possible values of k for which the curve with equation​​ 

y=-kx+k       k>0        x≠0 

does not touch or intersect the line with equation y = 3x + 4​​ 

(5)

SOLUTION​​ 

a-​​ Sketch the curve of negative reciprocal graph, with asymptotes x=0 and y=0.

b-​​ 

The graph cuts at​​ x-axis, so the y=0. So substituting the value of y in the equation of the curve.

y=-kx+k

0=-kx+k

kx=k

xk=x

x=1  

Hence, the coordinate of intersection of graph at x-axis is​​ (1,0).

c-​​ 

Now, solving both equations simultaneously by subtracting them.​​ 

3x+4= -kx+k

3x2+4x =-k+kx

3x2+4x-kx+k=0

3x2+4-kx+k=0

As it is given in the question, that​​ we need to find the possible range of values for which the curve doesn’t​​ intersect line, meaning thereby there is no solution. Therefore,​​ b2-4ac<0, where,​​ a=3,​​ b=4-k, and​​ c=k.

b2-4ac<0

4-k2-43k<0

16-8k +k2-12k<0

k2-20k+16<0

Let’s solve the quadratic equation to find the roots.

k2-20k+16=0

k-102-100= -16

k-102= - 16+100

k-102=84

k-10= ±84

k=10±84

k=10+84

k=  10-84

Since it was​​ k2-20k+16<0, therefore the​​ range of values of​​ x​​ would be​​ 

10-84<k<10+84