Question

Question #5ac74

Answer 1

First you must find the slope of your tangent line. This is equal to the derivative of your function `f(x)=5/x+1` evaluated in `x=4`:
`f'(x)=-5/x^2`
for `x=4`
`f'(4)=-5/16`

Point `x=4` corresponds to an `y` value on the function of `y=f(4)=9/4`
So the tangent line has slope `m=-5/16` and passes by the point on the curve of the function (`4,9/4`).
Insert these data into the general form for the line through a point and given slope: `y-y_0=m(x-x_0)`
You get:
`y-9/4=-5/16(x-4)`
And finally:
`y=-5/16x+7/2`

Graphically: