Start by finding the y-coordinate.
#f(3) = sqrt(3^2 - 4(3) + 7)/(3- 1) = sqrt(4)/2 = 1#
Differentiate using the quotient rule. Let #f(x) = (g(x))/(h(x))#, with #g(x) = sqrt(x^2 - 4x + 7)# and # h(x) = x - 1#.
We use the chain rule to differentiate #g(x)#:
Let #y = u^(1/2)# and #u = x^2 - 4x + 7#. Then #dy/(du) = 1/(2u^(1/2))# and #(du)/dx = 2x - 4#.
#g'(x)= dy/(du) * (du)/dx#
#g'(x) = 1/(2u^(1/2)) * 2x - 4#
#g'(x) = (2x- 4)/(2sqrt(x^2 - 4x + 7))#
#g'(x) = (2(x - 2))/(2sqrt(x^2 - 4x + 7))#
#g'(x)= (x- 2)/sqrt(x^2 - 4x + 7)#
Now use the quotient rule:
#f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2#
#f'(x) = ((x - 2)/sqrt(x^2 - 4x + 7)(x - 1) - sqrt(x^2 - 4x + 7)(1))/(x - 1)^2#
#f'(x) = ((x^2 - 2x - x + 2)/sqrt(x^2 - 4x + 7) - sqrt(x^2 - 4x + 7))/(x- 1)^2#
#f'(x) = ((x^2 - 3x + 2 - (x^2 - 4x + 7))/sqrt(x^2 - 4x + 7))/(x - 1)^2#
#f'(x) = (x - 5)/(sqrt(x^2 - 4x + 7)(x- 1)^2#
The slope of the tangent line can be obtained by evaluating the point #x = a# into the derivative.
#f'(3) = (3 - 5)/(sqrt(3^2 - 4(3) + 7)(3 - 1)^2#
#f'(3) = -2/(2(4))#
#f'(3) = -1/4#
The equation of the tangent is therefore:
#y - y_1 = m(x - x_1)#
#y- 1 = -1/4(x - 3)#
#y - 1 = -1/4x + 3/4#
#y = -1/4x + 7/4#
Hopefully this helps!
