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(A) Given the curve $y = x^3$. Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = 3x^2$. According to the problem,the slope of the tang

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$(3, 27)$

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(A) Given the curve $y = x^3$. Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = 3x^2$. According to the problem,the slope of the tangent at a point $(x, y)$ is equal to the $y$-coordinate of that point. So,$\frac{dy}{dx} = y$. Substituting the values,we have $3x^2 = x^3$. This implies $x^3 - 3x^2 = 0$,which factors to $x^2(x - 3) = 0$. Thus,$x = 0$ or $x = 3$. If $x = 0$,then $y = 0^3 = 0$. The point is $(0, 0)$. If $x = 3$,then $y = 3^3 = 27$. The point is $(3, 27)$. Comparing with the given options,$(3, 27)$ is the correct point.

If the slope of the tangent to the curve $y = x^3$ at a point is equal to the $y$-coordinate of that point,then the point is:

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