The curve y=x^3-2x^2+3x-4 intersects the hori | vedclass

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(B) Given the curve $y=x^3-2x^2+3x-4$ and the line $y=-2$. At the point of intersection $P(h, k)$,we have $x^3-2x^2+3x-4 = -2$. $\Rightarrow x^3-2x^2+

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$2$

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(B) Given the curve $y=x^3-2x^2+3x-4$ and the line $y=-2$. At the point of intersection $P(h, k)$,we have $x^3-2x^2+3x-4 = -2$. $\Rightarrow x^3-2x^2+3x-2 = 0$. By testing values,for $x=1$,$1-2+3-2 = 0$. Thus,the point $P$ is $(1, -2)$. Now,find the slope of the tangent at $P(1, -2)$: $\frac{dy}{dx} = 3x^2-4x+3$. At $x=1$,$\frac{dy}{dx} = 3(1)^2-4(1)+3 = 3-4+3 = 2$. The equation of the tangent at $(1, -2)$ is $y - (-2) = 2(x - 1)$. $\Rightarrow y+2 = 2x-2 \Rightarrow y = 2x-4$. This tangent meets the $X$-axis where $y=0$: $0 = 2x_1 - 4 \Rightarrow 2x_1 = 4 \Rightarrow x_1 = 2$.

The curve $y=x^3-2x^2+3x-4$ intersects the horizontal line $y=-2$ at the point $P(h, k)$. If the tangent drawn to this curve at $P$ meets the $X$-axis at $(x_1, y_1)$,then $x_1=$

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