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Question-12903




Question Number 12903 by tawa last updated on 06/May/17
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17
y^′ =6x−1⇒m_t =6×1−1=5  y−y_t =m_t (x−x_t )⇒y−2=5(x−1)  y=0⇒x−1=((−2)/5)⇒x=(3/5)⇒P((3/5),0) .■
$$\boldsymbol{{y}}^{'} =\mathrm{6}\boldsymbol{{x}}−\mathrm{1}\Rightarrow\boldsymbol{{m}}_{\boldsymbol{{t}}} =\mathrm{6}×\mathrm{1}−\mathrm{1}=\mathrm{5} \\ $$$$\boldsymbol{{y}}−\boldsymbol{{y}}_{\boldsymbol{{t}}} =\boldsymbol{{m}}_{\boldsymbol{{t}}} \left(\boldsymbol{{x}}−\boldsymbol{{x}}_{\boldsymbol{{t}}} \right)\Rightarrow\boldsymbol{{y}}−\mathrm{2}=\mathrm{5}\left(\boldsymbol{{x}}−\mathrm{1}\right) \\ $$$$\boldsymbol{{y}}=\mathrm{0}\Rightarrow\boldsymbol{{x}}−\mathrm{1}=\frac{−\mathrm{2}}{\mathrm{5}}\Rightarrow\boldsymbol{{x}}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow\boldsymbol{{P}}\left(\frac{\mathrm{3}}{\mathrm{5}},\mathrm{0}\right)\:.\blacksquare \\ $$
Commented by tawa last updated on 06/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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