The slope of the normal drawn at a point P to | vedclass

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(C) Given the curve $y = x^3 - 10x^2 + 31x - 30$. First, find the derivative: $\frac{dy}{dx} = 3x^2 - 20x + 31$. The slope of the normal is given by $

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$\frac{11}{7}$

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(C) Given the curve $y = x^3 - 10x^2 + 31x - 30$. First, find the derivative: $\frac{dy}{dx} = 3x^2 - 20x + 31$. The slope of the normal is given by $-\frac{1}{\frac{dy}{dx}}$. Setting this equal to $-\frac{1}{14}$: $-\frac{1}{3x^2 - 20x + 31} = -\frac{1}{14} \implies 3x^2 - 20x + 31 = 14$. $3x^2 - 20x + 17 = 0$. Factoring the quadratic: $(3x - 17)(x - 1) = 0$. This gives $x = 1$ or $x = \frac{17}{3}$. Since the problem states $x$ is an integer, we take $x = 1$. Substituting $x = 1$ into the curve equation: $y = (1)^3 - 10(1)^2 + 31(1) - 30 = 1 - 10 + 31 - 30 = -8$. So, the point $P$ is $(1, -8)$. The slope of the tangent at $x = 1$ is $\frac{dy}{dx} = 3(1)^2 - 20(1) + 31 = 14$. The equation of the tangent line is $y - (-8) = 14(x - 1) \implies y + 8 = 14x - 14 \implies y = 14x - 22$. For the $x$-intercept, set $y = 0$: $0 = 14x - 22 \implies 14x = 22 \implies x = \frac{22}{14} = \frac{11}{7}$.

The slope of the normal drawn at a point $P$ to the curve $y = x^3 - 10x^2 + 31x - 30$ is $-\frac{1}{14}$. Find the $x$-intercept of the tangent at point $P$, given that the $x$-coordinate of $P$ is an integer.

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