The tangent to the curve y=x^3+ax-b at the po | vedclass

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(A) Given the curve $y=x^3+ax-b$. The slope of the tangent is $\frac{dy}{dx} = 3x^2+a$. The slope of the line $y-x+4=0$ is $m_1 = 1$. Since the tangen

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$(2,-2)$

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(A) Given the curve $y=x^3+ax-b$. The slope of the tangent is $\frac{dy}{dx} = 3x^2+a$. The slope of the line $y-x+4=0$ is $m_1 = 1$. Since the tangent is perpendicular to the line,the slope of the tangent $m_2$ must satisfy $m_1 	imes m_2 = -1$,so $m_2 = -1$. At the point $(1,-5)$,$\frac{dy}{dx} = 3(1)^2+a = 3+a$. Setting $3+a = -1$,we get $a = -4$. Since the point $(1,-5)$ lies on the curve,we substitute $x=1, y=-5, a=-4$ into the equation: $-5 = (1)^3 + (-4)(1) - b -5 = 1 - 4 - b -5 = -3 - b b = 2$. The equation of the curve is $y = x^3 - 4x - 2$. Checking the options: For $(2,-2)$: $y = (2)^3 - 4(2) - 2 = 8 - 8 - 2 = -2$. Since the point $(2,-2)$ satisfies the equation,it lies on the curve.

The tangent to the curve $y=x^3+ax-b$ at the point $(1,-5)$ is perpendicular to the line $y-x+4=0$. Which one of the following points lies on the curve?

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