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

Apr 13, 2017

Any line through the point $\left(0 , 1\right)$ will have the form:
$y = m x + 1$
Substitute, the slope of the tangent line, ${m}_{t} = y ' \left(0\right)$ or the slope of the normal line, ${m}_{n} = - \frac{1}{y ' \left(0\right)}$

#### Explanation:

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

$y = m x + 1$

Compute $y ' \left(x\right)$:

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

Evaluate at the x coordinate:

${m}_{t} = y ' \left(0\right) = \frac{2}{3} {\left(0\right)}^{- \frac{1}{3}}$

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 $\left(0 , 1\right)$

Therefore, the tangent is the line, $x = 0$

The procedure for the normal line is similar.

$y = m x + 1$

Compute $y ' \left(x\right)$:

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

Evaluate at the x coordinate:

${m}_{n} = - \frac{1}{y ' \left(0\right)} = - \frac{1}{\left(\frac{2}{3}\right) {0}^{- \frac{1}{3}}} = 0$

Substitute 0 for the slope:

$y = 1$