Question

How do you find the equation of a line tangent to the function `y=2-sqrtx` at (4,0)?

Answer 1

`y=(-1/4)x+1`

Explanation:

The `color(red)(slope)` of the tangent line to the given function `2-sqrtx` is `color(red)(f'(4))`

Let us compute `color(red)(f'(4))`
`f(x)=2-sqrtx`
`f'(x)=0-1/(2sqrtx)=-1/(2sqrtx)`

`color(red)(f'(4))=-1/(2sqrt4)=-1/(2*2)=color(red)(-1/4)`

Since this line is tangent to the curve at `(color(blue)(4,0))`
then it passes through this point :
Equation of the line is :

`y-color(blue)0=color(red)(-1/4)(x-color(blue)4)`
`y=(-1/4)x+1`