N2008 P1 Q9

N2008 P1 Q9

Junior College 2
11 marks

It is given that \(f(x)=\frac{ax+b}{cx+d}\), for non-zero constants \(a,b,c,d\).

  1. Given that \(ad-bc\ne0\), show by differentiation that the graph of \(y=f(x)\) has no turning points.[3]
  2. What can be said about the graph of \(y=f(x)\) when \(ad-bc=0\)?[2]
  3. Deduce from part (i) that the graph of \(y=\frac{3x-7}{2x+1}\) has a positive gradient at all points of the graph.[1]
  4. On separate diagrams, draw sketches of the graphs of[5]
    1. \(y=\frac{3x-7}{2x+1}\),
    2. \(y^2=\frac{3x-7}{2x+1}\),

including the coordinates of the points where the graphs cross the axes and the equations of any asymptotes.

Solution:

Solution locked

Sign in to view the step-by-step solution

Finding similar questions...
Answer:(i) \(f'=(ad-bc)/(cx+d)^2\). (ii) \(y=a/c,x\ne-d/c\). (iv)(a) Intercepts \((7/3,0),(0,-7)\); asymptotes \(x=-1/2,y=3/2\). (b) \((7/3,0)\); asymptotes \(x=-1/2,y=\pm\sqrt{3/2}\).

Need help? Join our JC Math tuition classes.

Learn more