.
color(red)(a))a)
y=2/(2-x)y=22−x
P(3,-2)P(3,−2), Q(x, 2/(2-x))Q(x,22−x)
The slope of line PQ can be calculated for each xx-value using:
m_(PQ)=(y_Q-y_P)/(x_Q-x_P)mPQ=yQ−yPxQ−xP
x=2.9, :. y=2/(2-2.9)=2/(-0.9)=-2.222222, :. m_(PQ)=(-2.222222-(-2))/(2.9-3)=(-0.222222)/(-(0.1))=color(red)(2.222222)
x=2.99, :. y=2/(2-2.99)=2/(-0.99)=-2.020202, m_(PQ)=(-2.020202-(-2))/(2.99-3)=(-0.020202)/(-0.01)=color(red)(2.020202)
Much the same way:
x=2.999, :. y=-2.002002, m_(PO)=color(red)(2.002002)
x=2.9999, :. y=-2.00020002, m_(PO)=color(red)(2.000200)
x=3.1, :. y=-1.818181, m_(PQ)=color(red)(1.818181)
x=3.01, :. y=-1.980198, m_(PQ)=color(red)(1.980198)
x=3.001, :. y=-1.998001, m_(PQ)=color(red)(1.998001)
x=3.0001, :. y=-1.999800, m_(PQ)=color(red)(1.999800)
color(red)(b))
As we can see from part a), the values of the slope m converge from both directions to 2. Therefore,
color(red)(m_(PQ)=2)
color(red)(c))
The equation of the tangent line to the curve is:
y=mx+b where m is the slope and b is the y-intercept:
y=2x+b
We can use the coordinates of point P to solve for b:
-2=2(3)+b
-2=6+b
b=-8
Therefore, the equation of the tangent line is:
color(red)(y=2x-8)
The graph below shows the function and the tangent to it at point P: