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(B) The curve is $f(x) = x^2 - 4x + 6 = (x-2)^2 + 2$. This is a parabola with vertex at $(2,2)$.Let the line $L$ passing through $(2,3)$ have slope $m

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$y = 2$

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(B) The curve is $f(x) = x^2 - 4x + 6 = (x-2)^2 + 2$. This is a parabola with vertex at $(2,2)$.Let the line $L$ passing through $(2,3)$ have slope $m$. Its equation is $y - 3 = m(x - 2)$,or $y = mx - 2m + 3$.The area bounded by the line and the parabola is minimized when the line is parallel to the tangent at the point where the chord is bisected. For a parabola,the chord passing through a point $(h, k)$ that bisects the area is the one where the slope of the chord is equal to the slope of the tangent at the point $(h, k)$ if the point lies on the curve,but here $(2,3)$ is inside the parabola.Actually,for a parabola $y = ax^2 + bx + c$,the chord through a point $(x_0, y_0)$ that cuts off the minimum area is the one where the slope $m = f'(x_0)$.Here $f'(x) = 2x - 4$. At $x = 2$,$f'(2) = 2(2) - 4 = 0$.Thus,the slope of the line $L$ is $0$. The line $L$ is $y = 3$.The tangent parallel to $L$ (slope $0$) is the horizontal tangent at the vertex $(2,2)$,which is $y = 2$.

$A$ straight line $L$ passing through the point $(2,3)$ bounds the minimum area with the curve $f(x) = x^2 - 4x + 6$. Find the equation of the tangent to the curve parallel to $L$.

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