The curves y=x^2-1 and y=8x-x^2-9: | vedclass

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(B) To find the point of intersection,set the equations equal to each other:$x^2-1 = 8x-x^2-9$$2x^2-8x+8 = 0$$x^2-4x+4 = 0$$(x-2)^2 = 0$$x = 2$Substit

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touch each other at $(2,3)$

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(B) To find the point of intersection,set the equations equal to each other:$x^2-1 = 8x-x^2-9$$2x^2-8x+8 = 0$$x^2-4x+4 = 0$$(x-2)^2 = 0$$x = 2$Substituting $x=2$ into $y=x^2-1$,we get $y=3$. So,the point of intersection is $(2,3)$.Now,find the derivatives to determine the slopes of the tangents at $(2,3)$:For $C_1: y=x^2-1$,$\frac{dy}{dx} = 2x$. At $x=2$,$m_1 = 2(2) = 4$.For $C_2: y=8x-x^2-9$,$\frac{dy}{dx} = 8-2x$. At $x=2$,$m_2 = 8-2(2) = 4$.Since the slopes are equal $(m_1 = m_2 = 4)$,the tangents to the curves at the point $(2,3)$ are identical.Therefore,the curves touch each other at $(2,3)$.

The curves $y=x^2-1$ and $y=8x-x^2-9$:

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