Question

How do you find the equation of the line tangent to `f(x)= (sqrtx+1)`, at (0,1)?

Answer 1

Find the derivative at the point (= the slope of the graph's tangent). Then find the equation with the slope and point that you have.

Explanation:

• First: the slope=the derivative= `d/dx[f(x)]`
  `d/dx[sqrt(x)+1]=d/dx[x^(1/2)+1]`

`=1/2*x^(-1/2)` `=1/2*1/sqrt(x)`

`=1/(2sqrt(x))`

• The slope of the line tangent to f(x) at `(0,1)=`
  `1/(2sqrt(0))=1/0=>`the line is parallel to the `y`-axis

• Second: the equation of the tangent is
  `x=a` where `a` is constant
  again it passes through (0,1)
  so equation should be `x=0` i.e. `y` axis