# What is the normal to the tangent line at the y-intercept to y = 1/3x^3 - 4x + 2?

Dec 24, 2016

$y = \frac{1}{4} x + 2$

#### Explanation:

The y-intercept will occur at $\left(0 , 2\right)$:

$y = \frac{1}{3} {\left(0\right)}^{3} - 4 \left(0\right) + 2 = 2$

The function's derivative can be found by the power rule, which states that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$.

$y ' = {x}^{2} - 4$

The slope of the tangent is given by evaluating your point $x = a$ into the derivative. The normal line is perpendicular to the tangent line, or the product of their two slopes equals $- 1$.

The slope of the tangent is $m = {0}^{2} - 4 = 0 - 4 = - 4$. Then the slope of the normal line is $\frac{1}{4}$.

Since the reaction passes through $\left(0 , 2\right)$, the normal line has equation:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 2 = \frac{1}{4} \left(x - 0\right)$

$y - 2 = \frac{1}{4} x$

$y = \frac{1}{4} x + 2$

Here is a graphical depiction of the problem. The graph in $\textcolor{red}{\text{red}}$ is the function $y = \frac{1}{3} {x}^{3} - 4 x + 2$, the graph in $\textcolor{p u r p \le}{\text{purple}}$ is the tangent line and the graph in $\textcolor{b l u e}{\text{blue}}$ is the normal line.

Hopefully this helps!