The slope of a curve passing through the poin | vedclass

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(C) Given the slope of the curve is $\frac{dy}{dx} = x^2 - 2x$.Integrating both sides with respect to $x$:$y = \int (x^2 - 2x) dx = \frac{x^3}{3} - x^

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$(0, 4/3)$

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(C) Given the slope of the curve is $\frac{dy}{dx} = x^2 - 2x$.Integrating both sides with respect to $x$:$y = \int (x^2 - 2x) dx = \frac{x^3}{3} - x^2 + C$.Since the curve passes through the point $(2, 0)$,we substitute $x = 2$ and $y = 0$ into the equation:$0 = \frac{2^3}{3} - 2^2 + C$$0 = \frac{8}{3} - 4 + C$$0 = \frac{8 - 12}{3} + C$$0 = -\frac{4}{3} + C \Rightarrow C = \frac{4}{3}$.Thus,the equation of the curve is $y = \frac{x^3}{3} - x^2 + \frac{4}{3}$.To find the value of $y$ at $x = 0$:$y(0) = \frac{0^3}{3} - 0^2 + \frac{4}{3} = \frac{4}{3}$.Therefore,the point is $(0, 4/3)$.

The slope of a curve passing through the point $(2, 0)$ at $(x, y)$ is $x^2 - 2x$. Find the value of $y$ at $x = 0$.

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