Find the equation of the tangent to the curve | vedclass

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(A) On the $x$-axis,the value of $y = 0$. Setting $y = 0$ in the curve equation $\frac{x-7}{(x-2)(x-3)} = 0$,we get $x = 7$. Thus,the point of interse

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$20y - x + 7 = 0$

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(A) On the $x$-axis,the value of $y = 0$. Setting $y = 0$ in the curve equation $\frac{x-7}{(x-2)(x-3)} = 0$,we get $x = 7$. Thus,the point of intersection is $(7, 0)$.To find the slope of the tangent,we differentiate $y = \frac{x-7}{x^2 - 5x + 6}$ with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(x^2 - 5x + 6)(1) - (x-7)(2x-5)}{(x^2 - 5x + 6)^2}$.At the point $(7, 0)$,the denominator is $(7^2 - 5(7) + 6)^2 = (49 - 35 + 6)^2 = (20)^2 = 400$.The numerator at $x=7$ is $(49 - 35 + 6) - (7-7)(2(7)-5) = 20 - 0 = 20$.Thus,the slope $m = \frac{dy}{dx} \Big|_{(7,0)} = \frac{20}{400} = \frac{1}{20}$.The equation of the tangent line at $(7, 0)$ with slope $m = \frac{1}{20}$ is given by $y - y_1 = m(x - x_1)$:$y - 0 = \frac{1}{20}(x - 7)$,$20y = x - 7$,$20y - x + 7 = 0$.

Find the equation of the tangent to the curve $y = \frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the $x$-axis.

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