find the x coordinate of the point on the curve y=x²-3x+1 where the tangent is parallel to the line y=x

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$$\begin{gathered} \mathrm{Slope} \mathrm{of} \mathrm{line} \mathrm{y}=\mathrm{x} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{m}=1 \left[\mathrm{Compare} \mathrm{with} \mathrm{y}=\mathrm{mx}\right] \\ \mathrm{As} \mathrm{tangent} \mathrm{to} \mathrm{curve} \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{y}=\mathrm{x}, \mathrm{hence} \\ \mathrm{Slope} \mathrm{of} \mathrm{tangent} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{m}=1 \\ \mathrm{Let} \mathrm{requred} \mathrm{point} \mathrm{be} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \\ \mathrm{Now} \\ \mathrm{y}={\mathrm{x}}^2-3\mathrm{x}+1 \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}, \mathrm{we} \mathrm{get}: \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2-3\mathrm{x}+1\right) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}-3 \\ \Rightarrow {\frac{\mathrm{dy}}{\mathrm{dx}}}_{\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)}=2{\mathrm{x}}_1-3 \\ \mathrm{But} {\frac{\mathrm{dy}}{\mathrm{dx}}}_{\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)}=\mathrm{Slope} \mathrm{of} \mathrm{tangent}=\mathrm{m}=1 \\ \Rightarrow 2{\mathrm{x}}_1-3=1 \\ \Rightarrow 2{\mathrm{x}}_1=3+1 \\ \Rightarrow 2{\mathrm{x}}_1=4 \\ \mathbf{\Rightarrow }{\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{ }\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$
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