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

Question

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

Accepted Answer

(i) $\mathrm f'(x)=\dfrac{ad-bc}{(cx+d)^2}$, so there are no turning points when $ad-bc\neq0$ (ii) $y=\dfrac ac$ with a hole at $\left(-\dfrac dc,\dfrac ac\right)$ (iii) $\mathrm f'(x)=\dfrac{17}{(2x+1)^2}>0$ (iv) Intercepts $\left(\dfrac73,0\right)$, $(0,-7)$, asymptotes $x=-\dfrac12$, $y=\dfrac32$

N2008 P1 Q9 [Modified]

N2008 P1 Q9 [Modified]

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