How do you find the equation of the tangent and normal line to the curve y=1+x23 at (0,1)?

Question

xxkrnjangxxed9q · Accepted Answer

Usually we would use the form given in the above answer:
y=mx+1
Compute y'(x):

y^{'} (x) = \frac{2}{3} x^{- \frac{1}{3}}

Evaluate at the x coordinate:

m_{t} = y^{'} (0) = \frac{2}{3} {(0)}^{- \frac{1}{3}}

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 (0,1)
Therefore, the tangent is the line, x=0
The procedure for the normal line is similar.
y=mx+1
Compute y'(x):

y^{'} (x) = \frac{2}{3} x^{- \frac{1}{3}}

Evaluate at the x coordinate:

m_{n} = - \frac{1}{y^{'} (0)} = - \frac{1}{(\frac{2}{3}) 0^{- \frac{1}{3}}} = 0

Substitute 0 for the slope:
y=1

How do you find the equation of the

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