Equation of the normal to the curve y=x^2+x a | vedclass

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(D) Given curve is $y=x^2+x$. First,find the derivative $\frac{dy}{dx} = 2x+1$. The slope of the tangent at point $(1,2)$ is $\left. \frac{dy}{dx} \ri

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$x+3y-7=0$

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(D) Given curve is $y=x^2+x$. First,find the derivative $\frac{dy}{dx} = 2x+1$. The slope of the tangent at point $(1,2)$ is $\left. \frac{dy}{dx} ight|_{(1,2)} = 2(1)+1 = 3$. The slope of the normal is $m = \frac{-1}{	ext{slope of tangent}} = \frac{-1}{3}$. The equation of the normal at point $(x_1, y_1) = (1,2)$ is given by $y - y_1 = m(x - x_1)$. Substituting the values: $y - 2 = \frac{-1}{3}(x - 1)$. Multiplying by $3$: $3y - 6 = -x + 1$. Rearranging the terms: $x + 3y - 7 = 0$.

Equation of the normal to the curve $y=x^2+x$ at the point $(1,2)$ is

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