How do you find the equation of the line tangent to #y=(2x+1)/(x+2)#with the given point of (1,1)?

1 Answer
Jun 24, 2016

We must first differentiate using the quotient rule.

Explanation:

Let #y= (g(x))/(h(x))#

Then #g(x) = 2x + 1# and #h(x) = x + 2#. The quotient rule states that #y' = (g'(x) xx h(x) - (g(x) xx h'(x)))/(h(x)^2#

By the power rule:

#g'(x) = 2#

#h'(x) = 1#

We now have all the information we need to know:

#y' =(2(x + 2) - 1(2x + 1))/(x + 2)^2#

#y' = (2x + 4 - 2x - 1)/(x^2 + 4x + 4)#

#y' = 3/(x^2 + 4x + 4)#

The slope of the tangent is given by evaluating #f(a)# in the derivative function, #x = a# being the given point.

#y = 3/(1^2 + 4 + 4)#

#y = 1/3#

The slope of the tangent is #1/3#. Now we know the slope of the tangent and a point that lies both on the function and on the tangent.

By point-slope form:

#y - y_1 + m(x - x_1)#

#y - 1 = 1/3(x - 1)#

#y - 1 = 1/3x - 1/3#

#y = 1/3x + 2/3#

The equation of the tangent is #y = 1/3x + 2/3#.

Hopefully this helps!