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(B) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx} = 6x^2+10x-9$. Integrating both sides with respect to $x$, we get: $f(x)

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$4$

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(B) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx} = 6x^2+10x-9$. Integrating both sides with respect to $x$, we get: $f(x) = \int (6x^2+10x-9) dx = 2x^3+5x^2-9x+C$. Given that $f(2)=0$, we substitute $x=2$ into the equation: $f(2) = 2(2)^3 + 5(2)^2 - 9(2) + C = 0$ $16 + 20 - 18 + C = 0$ $18 + C = 0 \Rightarrow C = -18$. Thus, the function is $f(x) = 2x^3+5x^2-9x-18$. Now, we find $f(-2)$: $f(-2) = 2(-2)^3 + 5(-2)^2 - 9(-2) - 18$ $f(-2) = 2(-8) + 5(4) + 18 - 18$ $f(-2) = -16 + 20 + 0 = 4$.

If the slope of the tangent drawn at any point $(x, y)$ on the curve $y=f(x)$ is $(6x^2+10x-9)$ and $f(2)=0$, then $f(-2)=$

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