Find the equations of the tangent and the nor | vedclass

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(A) Given the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = 4x^{3}-18x^{2}+26x-10$. At the poi

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Tangent: $2x-y+1=0$,Normal: $x+2y-7=0$

Vedclass · Answer

(A) Given the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = 4x^{3}-18x^{2}+26x-10$. At the point $(1,3)$,the slope of the tangent is: $m = \left. \frac{dy}{dx} \right|_{(1,3)} = 4(1)^{3}-18(1)^{2}+26(1)-10 = 4-18+26-10 = 2$. The equation of the tangent at $(1,3)$ is: $y-3 = 2(x-1) \Rightarrow y-3 = 2x-2 \Rightarrow 2x-y+1 = 0$. The slope of the normal at $(1,3)$ is: $m' = -\frac{1}{m} = -\frac{1}{2}$. The equation of the normal at $(1,3)$ is: $y-3 = -\frac{1}{2}(x-1) \Rightarrow 2y-6 = -x+1 \Rightarrow x+2y-7 = 0$.

Find the equations of the tangent and the normal to the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at the point $(1,3)$.

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