Question

solve

$$\begin{gathered} a)Findthepointsonthecurvey=x^3-10x+8atwhichthe\tan gentisparalleltotheliney=2x+1 \\ b)Isthegivenliney=2x+1\tan genttothecurve?Why? \end{gathered}$$

Answer 1

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$$\begin{gathered} \mathrm{Let}\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\mathrm{be}\mathrm{the}\mathrm{required}\mathrm{point}.\mathrm{The}\mathrm{given}\mathrm{curve}\mathrm{is} \\ \mathrm{y}={\mathrm{x}}^3-10\mathrm{x}+8...\left(1\right) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^2-10 \\ \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)}=3{{\mathrm{x}}_1}^2-10 \\ \mathrm{Since}\mathrm{the}\mathrm{tangent}\mathrm{at}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{y}=2\mathrm{x}+1 \\ \mathrm{So},\mathrm{slope}\mathrm{of}\mathrm{tangent}\mathrm{at}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{y}=2\mathrm{x}+1 \\ \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)}=2\left[\because \mathrm{y}=\mathrm{mx}+\mathrm{c}\mathrm{has}\mathrm{slope}\mathrm{m},\mathrm{so}\mathrm{y}=2\mathrm{x}+1\mathrm{has}\mathrm{slope}2\right] \\ \Rightarrow 3{{\mathrm{x}}_1}^2-10=2 \\ \Rightarrow 3{{\mathrm{x}}_1}^2=12 \\ \Rightarrow {{\mathrm{x}}_1}^2=4 \\ \Rightarrow {\mathrm{x}}_1=\pm 2 \\ \mathrm{Since}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}\left(1\right),\hspace{0.17em}\mathrm{So},{\mathrm{y}}_1={{\mathrm{x}}_1}^3-10{\mathrm{x}}_1+8 \\ \mathrm{Now},\mathrm{when}{\mathrm{x}}_1=2\Rightarrow {\mathrm{y}}_1=2^3-10\left(2\right)+8=-4 \\ \mathrm{When}{\mathrm{x}}_1=-2\Rightarrow {\mathrm{y}}_1={\left(-2\right)}^3-10\left(-2\right)+8=-8+20+8=20 \\ \mathrm{Thus}\mathrm{the}\mathrm{required}\mathrm{points}\mathrm{are}\left(2,-4\right)\mathrm{and}\left(-2,20\right) \end{gathered}$$
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