If the slope of the tangent drawn at any poin | vedclass

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(B) Given $\frac{dy}{dx} = 3x^2 - 5$. Integrating both sides, we get $y = \int (3x^2 - 5) dx = x^3 - 5x + C$. Since $f(1) = 2$, we have $2 = 1^3 - 5(1

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$(-2, 8)$

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(B) Given $\frac{dy}{dx} = 3x^2 - 5$. Integrating both sides, we get $y = \int (3x^2 - 5) dx = x^3 - 5x + C$. Since $f(1) = 2$, we have $2 = 1^3 - 5(1) + C$, which gives $2 = 1 - 5 + C$, so $C = 6$. Thus, the equation of the curve is $y = x^3 - 5x + 6$. The slope of the tangent at $(1, 2)$ is $\left. \frac{dy}{dx} \right|_{(1, 2)} = 3(1)^2 - 5 = -2$. The equation of the tangent at $(1, 2)$ is $y - 2 = -2(x - 1)$, which simplifies to $y = -2x + 4$. To find the intersection point, set the curve equation equal to the tangent equation: $x^3 - 5x + 6 = -2x + 4$. This simplifies to $x^3 - 3x + 2 = 0$. Factoring the cubic equation, we get $(x - 1)^2(x + 2) = 0$. The roots are $x = 1$ and $x = -2$. For $x = 1$, $y = 2$ (the point of tangency). For $x = -2$, $y = -2(-2) + 4 = 8$. Therefore, the tangent intersects the curve at the point $(-2, 8)$.

If the slope of the tangent drawn at any point $(x, y)$ to the curve $y=f(x)$ is $3x^2-5$ and $f(1)=2$, then the tangent at $(1, 2)$ to the curve $y=f(x)$ intersects the curve at the point

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