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Question
Find the point on the curve y = x2 − 2x + 3, where the tangent is parallel to x-axis ?
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Solution
The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.
Here,
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence,} y_1 = {x_1}^2 - 2 x_1 + 3 . . . \left( 1 \right)\]
\[\text { Now }, y = x^2 - 2x + 3\]
\[\frac{dy}{dx} = 2x - 2\]
\[\text { Slope of the tangent at }\left( x, y \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =2 x_1 - 2\]
\[\text { Given }:\]
\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)=\text { Slope of thex-axis }\]
\[ = 2 x_1 - 2 = 0\]
\[ \Rightarrow x_1 = 1\]
\[\text { and }\]
\[ y_1 = 1 - 2 + 3 = 2 [\text { From } (1)]\]
\[ \therefore \text { Required point }=\left( x_1 , y_1 \right)=\left( 1, 2 \right)\]
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