If the tangent to the graph of the function #y = (x- 1)/x# passes through the point #(4, 1)#, then what is the equation of the tangent line?
1 Answer
The equation is
Explanation:
We want the graph to pass through
#f(x) = 1 - 1/x#
#f'(x) = 1/x^2#
Now
#y - y_1 = m(x -x_1)#
#y - 1 = 1/x^2(x - 4)#
#y = 1/x - 4/x^2 + 1#
We will need a second equation. Because the point
#y = 1 - 1/x# .
We can immediately solve through substitution.
#1 - 1/x = 1/x - 4/x^2 +1#
#0 = 2/x - 4/x^2#
#0 = (2x - 4)/x^2#
#0 = 2x - 4#
#4 = 2x#
#x = 2#
Therefore, the slope of the line will be
#y - 1 = 1/4(x - 4)#
#y - 1 = 1/4x - 1#
#y = 1/4x#
Let's confirm graphically.
The point in orange is the point of tangency, the point in green is the point
This proves that our answer is viable.
Hopefully this helps!