Question Number 187053 by horsebrand11 last updated on 13/Feb/23
$$\:{How}\:{do}\:{you}\:{make}\:{a}\:{curve}\: \\ $$$$\:{y}={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}\:{with}\:{a}\:{critical} \\ $$$$\:{point}\:{of}\:\left(\mathrm{1},\mathrm{0}\right)\:{and}\:\left(−\mathrm{2},\mathrm{27}\right)\:? \\ $$
Answered by cortano12 last updated on 13/Feb/23
$$\:\Rightarrow\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}\:=\:{k}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}\:=\:{kx}^{\mathrm{2}} +{kx}−\mathrm{2}{k} \\ $$$$\Rightarrow\begin{cases}{{k}=\mathrm{3}{a}=\mathrm{2}{b}\:;\:{b}=\frac{\mathrm{3}}{\mathrm{2}}{a}}\\{{c}=−\mathrm{2}{k}=−\mathrm{6}{a}}\end{cases} \\ $$$$\because\:{y}={ax}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{ax}^{\mathrm{2}} −\mathrm{6}{ax}+{d} \\ $$$$\:\Rightarrow−\frac{\mathrm{7}}{\mathrm{2}}{a}+{d}=\mathrm{0}\Rightarrow{d}=\frac{\mathrm{7}}{\mathrm{2}}{a} \\ $$$$\Rightarrow\mathrm{10}{a}+{d}=\mathrm{27}\:\Rightarrow\frac{\mathrm{27}}{\mathrm{2}}{a}=\mathrm{27}\:;\:{a}=\mathrm{2} \\ $$$$\:\begin{cases}{{d}=\mathrm{7}}\\{{c}=−\mathrm{12}}\\{{b}=\mathrm{3}}\end{cases}\:\Rightarrow{y}=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}\: \\ $$