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

Question

It is given that $f(x)=\frac{ax+b}{cx+d}$, for non-zero constants $a,b,c,d$. Given that $ad-bc\ne0$, show by differentiation that the graph of $y=f(x)$ has no turning points.[3] What can be said about the graph of $y=f(x)$ when $ad-bc=0$?[2] 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] On separate diagrams, draw sketches of the graphs of[5] $y=\frac{3x-7}{2x+1}$, $y^2=\frac{3x-7}{2x+1}$, including the coordinates of the points where the graphs cross the axes and the equations of any asymptotes.

Accepted 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 $x=-1/2,y=\pm\sqrt{3/2}$.

N2008 P1 Q9

N2008 P1 Q9

Solution:

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