If (2, a) and (b, 19) are two stationary poin | vedclass

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(B) Given the curve $y=2x^3-15x^2+36x+c$. Stationary points occur where $\frac{dy}{dx} = 0$. $\frac{dy}{dx} = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-

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$15$

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(B) Given the curve $y=2x^3-15x^2+36x+c$. Stationary points occur where $\frac{dy}{dx} = 0$. $\frac{dy}{dx} = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$. Setting $\frac{dy}{dx} = 0$,we get $x=2$ and $x=3$. Since $(2, a)$ is a point on the curve,$a = 2(2)^3 - 15(2)^2 + 36(2) + c = 16 - 60 + 72 + c = 28 + c$. Since $(b, 19)$ is a point on the curve,$b$ must be the other $x$-coordinate,so $b=3$. Substituting $x=3$ and $y=19$ into the curve equation: $19 = 2(3)^3 - 15(3)^2 + 36(3) + c$. $19 = 54 - 135 + 108 + c \Rightarrow 19 = 27 + c \Rightarrow c = -8$. Now,$a = 28 + (-8) = 20$. Therefore,$a+b+c = 20 + 3 - 8 = 15$.

If $(2, a)$ and $(b, 19)$ are two stationary points of the curve $y=2x^3-15x^2+36x+c$,then $a+b+c=$

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