The equation of the tangent to the curve $y = \frac{(x - 3)}{(x - 1)(x - 2)}$ at the point, where it cuts x-axis is:

When a curve "cuts the x-axis," it crosses the x-axis where $y = 0$.

Set $y = 0$:

$0 = \frac{(x - 3)}{(x - 1)(x - 2)}$

A fraction equals zero only when its numerator is zero (and denominator is not zero).

$x - 3 = 0$

$x = 3$

The curve cuts the x-axis at the point $(3, 0)$.

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To find the slope of the tangent at $(3, 0)$, find the derivative $\frac{dy}{dx}$.

Given: $y = \frac{(x - 3)}{(x - 1)(x - 2)}$

Using the Quotient Rule: $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$

Where:

• $u = (x - 3)$, so $u' = 1$
• $v = (x - 1)(x - 2) = x^2 - 3x + 2$, so $v' = 2x - 3$

$\frac{dy}{dx} = \frac{1 \cdot (x^2 - 3x + 2) - (x - 3)(2x - 3)}{[(x - 1)(x - 2)]^2}$

Simplifying the numerator:

$(x^2 - 3x + 2) - (x - 3)(2x - 3)$

$= x^2 - 3x + 2 - (2x^2 - 3x - 6x + 9)$

$= x^2 - 3x + 2 - 2x^2 + 9x - 9$

$= -x^2 + 6x - 7$

At $x = 3$:

Numerator: $-(3)^2 + 6(3) - 7 = -9 + 18 - 7 = 2$

Denominator: $[(3-1)(3-2)]^2 = [2 \times 1]^2 = 4$

Slope $= \frac{2}{4} = \frac{1}{2}$

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Using point-slope form with point $(3, 0)$ and slope $m = \frac{1}{2}$:

$y - 0 = \frac{1}{2}(x - 3)$

$y = \frac{1}{2}x - \frac{3}{2}$

$2y = x - 3$

$2y - x + 3 = 0$

Therefore, the equation of the tangent is $2y - x + 3 = 0$.