Determine an equation of a tangent line to the curve y=-3x+2/x+100 at the point where x=2? thanks

1 Answer
Feb 27, 2018

y=-151/5202x-253/2601

Explanation:

1. First, find the derivative of the original equation using the quotient rule.

y=(-3x+2)/(x+100)

This is the quotient rule:

d/dx[f(x)/g(x)]=(f'(x)g(x)-f(x)g'(x))/[g(x)]^2

Plug the values in:

dy/dx=((-3x+2)'(x+100)-(-3x+2)(x+100)')/(x+100)^2

Differentiate the individual terms:

dy/dx=(-3(x+100)-1(-3x+2))/(x+100)^2

Simplify:

dy/dx=(cancel(color(red)(-3x))-300cancel(color(red)(+3x))-2)/(x^2+200x+10000)

dy/dx=-302/(x^2+200x+10000)

2. The derivative of a function at a certain point gives the slope of the line tangent to that point.

First, we will equat f(x) and y to simply the notation.

f(x)=y

Next, find the y-value corresponding to x=2 by plugging x=2 into the original function.

f(2)=(-3(2)+2)/((2)+100)=-4/102=-2/51

So, the point we want to find the equation of the line tangent to is (2,-2/51).

Now, we can plug x=2 into our derivative to find the slope of the line tangent to that point.

f'(2)=-302/((2)^2+200(2)+10000)

Simplify:

f'(x)=-302/(4+400+10000)=-151/5202

Finally, plug this information into the point-slope formula:

y_2-y_1=m(x_2-x_1)

y+2/51=-151/5202(x-2)

This can also be written in slope intercept form if y is isolated:

y+2/51=-151/5202x-151/2601

y=-151/5202x-253/2601