Let #f(x) = (x+2)/(x+3)#. Find the equation(s) of tangent line(s) that pass through a point (0,6)? Sketch the solution?

Question

Let #f(x) = (x+2)/(x+3)#. Find the equation(s) of tangent line(s) that pass through a point (0,6)?

Emily Ammerman · Accepted Answer

Tangents are #25x-9y+54=0# and #y=x+6#

Let the slope of the tangent be #m#. The equation of tangent then is #y-6=mx# or #y=mx+6#

Now let us see the point of intersection of this tangent and given curve #y=(x+2)/(x+3)#. For this putting #y=mx+6# in this we get

#mx+6=(x+2)/(x+3)# or #(mx+6)(x+3)=x+2#

i.e. #mx^2+3mx+6x+18=x+2#

or #mx^2+(3m+5)x+16=0#

This should give two values of #x# i.e. two points of intersection, but tangent cuts the curve only at one point. Hence if #y=mx+6# is a tangent, we should have only one root for the quadratic equation, which is possible onli if discriminant is #0# i.e.

#(3m+5)^2-4*m*16=0#

or #9m^2+30m+25-64m=0#

or #9m^2-34m+25=0#

i.e. #m=(34+-sqrt(34^2-900))/18#

= #(34+-sqrt256)/18=(34+-16)/18#

i.e. #25/9# or #1#

and hence tangents are #y=25/9x+6# i.e. #25x-9y+54=0#

and #y=x+6#

graph{(25x-9y+54)(x-y+6)(y-(x+2)/(x+3))=0 [-12.58, 7.42, -3.16, 6.84]}

Charles Arocha · Answer

undefined To find the equation(s) of tangent line(s) that pass through the point (0,6) on the graph of f(x) = (x+2)/(x+3), we need to find the derivative of f(x) and substitute the coordinates of the given point into the derivative equation.

The derivative of f(x) can be found using the quotient rule:

f'(x) = [(x+3)(1) - (x+2)(1)] / (x+3)^2

Simplifying this expression gives:

f'(x) = 1 / (x+3)^2

Now, substitute the coordinates of the point (0,6) into the derivative equation:

6 = 1 / (0+3)^2

Solving for the constant value gives:

6 = 1 / 9

Multiplying both sides by 9 gives:

54 = 1

This is not possible, so there is no tangent line passing through the point (0,6) on the graph of f(x) = (x+2)/(x+3).

Therefore, there are no tangent lines to sketch.