Find the equations of the tangent and normal | 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 point

Vedclass · Accepted Answer

Tangent: $10x+y-5=0$,Normal: $x-10y+50=0$

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 $(0,5)$,the slope of the tangent is:$m = \left. \frac{dy}{dx} \right|_{(0,5)} = 4(0)^{3}-18(0)^{2}+26(0)-10 = -10$.The equation of the tangent at $(0,5)$ is:$y - 5 = -10(x - 0) \Rightarrow y - 5 = -10x \Rightarrow 10x + y - 5 = 0$.The slope of the normal at $(0,5)$ is:$m' = -\frac{1}{m} = -\frac{1}{-10} = \frac{1}{10}$.The equation of the normal at $(0,5)$ is:$y - 5 = \frac{1}{10}(x - 0) \Rightarrow 10y - 50 = x \Rightarrow x - 10y + 50 = 0$.

Find the equations of the tangent and normal to the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at the point $(0,5)$.

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