Question

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

Answer 1

`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`