Question

find the equation of the tangent to the curve y = x-7/((x-2)(x-3)) at the point where it cuts the x-axis?

Answer 1

First u solve the( x-2)(x-3) equation then solve x-x-7

Answer 2

Answer:

20y-x+7 = 0.

Step-by-step explanation:

As the point cut at the x-axis, then y=0. Hence, the equation of the curve, if y=0, then the value of x is 7. (i.e., x=7). Hence, the curve cuts the x-axis at (7,0)

Now, differentiate the equation of the curve with respect to x, we get

dy/dx = [(1-y)(2x-5)] / [(x-2)((x-3)]

dy/dx](7, 0) = (1-0)/[(5)(4)] = 1/20

Hence, the slope of the tangent line at (7, 0) is 1/20.

Therefore, the equation of the tangent at (7, 0) is

Y-0 = (1/20)(x-7)

20y-x+7 = 0.

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