Question Number 120036 by I want to learn more last updated on 28/Oct/20
Answered by bemath last updated on 28/Oct/20
$$\Rightarrow{gradient}\:=\:\frac{{dy}}{{dx}} \\ $$$$\Rightarrow\mathrm{3}=\mathrm{4}{x}+\mathrm{5}\:;\:\begin{cases}{{x}=−\frac{\mathrm{1}}{\mathrm{2}}}\\{{y}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{1}=−\mathrm{1}}\end{cases} \\ $$$${Thus}\:{equation}\:{of}\:{tangent}\:{to}\:{curve} \\ $$$${y}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\:{is}\:\mathrm{3}{x}−{y}=−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1} \\ $$$$\mathrm{3}{x}−{y}=−\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\mathrm{6}{x}−\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$
Commented by I want to learn more last updated on 29/Oct/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Commented by I want to learn more last updated on 29/Oct/20
$$\mathrm{Sir},\:\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\:\:\:\:\mathrm{3x}\:\:−\:\:\mathrm{y}\:\:\:\:\:\:\mathrm{from}\:\:\:\:\:\mathrm{y}\:\:\:=\:\:\:\mathrm{2x}^{\mathrm{2}} \:\:+\:\:\mathrm{5x}\:\:+\:\:\mathrm{1} \\ $$