The equation of the tangent to the curve y=x^ | vedclass

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(A) Given the curve $y=x^3-2x+7$. To find the slope of the tangent,we differentiate with respect to $x$: $\frac{dy}{dx} = 3x^2-2$. At the point $(1,6)

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$y=x+5$

Vedclass · Answer

(A) Given the curve $y=x^3-2x+7$. To find the slope of the tangent,we differentiate with respect to $x$: $\frac{dy}{dx} = 3x^2-2$. At the point $(1,6)$,the slope $m$ is: $m = \left. \frac{dy}{dx} \right|_{x=1} = 3(1)^2-2 = 3-2 = 1$. The equation of the tangent line passing through $(x_1, y_1) = (1,6)$ with slope $m=1$ is given by: $y-y_1 = m(x-x_1)$ $y-6 = 1(x-1)$ $y-6 = x-1$ $y = x+5$. Thus,the correct option is $A$.

The equation of the tangent to the curve $y=x^3-2x+7$ at the point $(1,6)$ is

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