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Question
Find the equation of the tangent and the normal to the following curve at the indicated point y = x4 − 6x3 + 13x2 − 10x + 5 at x = 1?
Solution
\[y= x^4 - 6 x^3 + 13 x^2 - 10x + 5\]
\[\text{ When }x = 1 , \]
`y = 1 - 6 + 13 - 10 + 5 = 3`
\[\text { So}, \left( x_1 , y_1 \right) = \left( 1, 3 \right)\]
\[\text { Now,} y= x^4 - 6 x^3 + 13 x^2 - 10x + 5\]
\[\text { Differentiating both sides w.r.t.x,} \]
\[\frac{dy}{dx} = 4 x^3 - 18 x^2 + 26x - 10\]
\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( 1, 3 \right) =4-18+26 - 10 = 2\]
\[\text { Equation of tangent is },\]
\[y - y_1 = 2 \left( x - x_1 \right)\]
\[ \Rightarrow y - 3 = 2\left( x - 1 \right)\]
\[ \Rightarrow y - 3 = 2x - 2\]
\[ \Rightarrow 2x - y + 1 = 0\]
\[\text { Equation of normal is
},\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 3 = \frac{- 1}{2} \left( x - 1 \right)\]
\[ \Rightarrow 2y - 6 = - x + 1\]
\[ \Rightarrow x + 2y - 7 = 0\]
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