Question Number 156188 by naka3546 last updated on 09/Oct/21
$${Given}\:\:{a}\:\:{rational}\:\:{function} \\ $$$$\:\:\:\:\:{f}\left({x}\right)\:=\:\:\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{x}+{q}} \\ $$$${has}\:\:{minimum}\:\:{point}\:\:{at}\:\left(−\mathrm{2},\mathrm{9}\right)\:\:{and}\:\:{maximum}\:\:{point}\:\:{at}\:\:\left(\mathrm{2},\mathrm{1}\right)\:. \\ $$$${Find}\:\:{value}\:\:{of}\:\:{a},\:{b},\:{c},\:\:{and}\:\:{q}\:. \\ $$
Commented by MJS_new last updated on 09/Oct/21
$$\mathrm{we}\:\mathrm{have}\:\mathrm{4}\:\mathrm{unknowns}\:\mathrm{and}\:\mathrm{need}\:\mathrm{4}\:\mathrm{equations} \\ $$$$\left(\mathrm{1}\right)\:{f}\left(−\mathrm{2}\right)=\mathrm{9} \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\mathrm{2}\right)=\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{f}'\left(−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:{f}'\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{to}\:\mathrm{get} \\ $$$${a}=−\mathrm{1} \\ $$$${b}=\mathrm{5} \\ $$$${c}=−\mathrm{4} \\ $$$${q}=\mathrm{0} \\ $$
Commented by naka3546 last updated on 09/Oct/21
$${Thank}\:\:{you},\:{sir}. \\ $$