# How do you find the equation of the tangent and normal line to the curve #y=1+x^(2/3)# at (0,1)?

##### 1 Answer

Apr 13, 2017

Any line through the point

Substitute, the slope of the tangent line,

#### Explanation:

Usually we would use the form given in the above answer:

Compute

Evaluate at the x coordinate:

But we have an exception to the procedure. Please observe that that the evaluation causes a division by 0, which implies that the tangent is the vertical passing the point

Therefore, the tangent is the line,

The procedure for the normal line is similar.

Compute

Evaluate at the x coordinate:

Substitute 0 for the slope: