# What is the equation of the line normal to f(x)= x^3-e^x  at x=2?

$y = \frac{2 - x}{12 - {e}^{2}} - 8 + {e}^{2}$

#### Explanation:

First, we have to Find the Required Coordinate on which the Normal Line will intersect with the graph.

So, We are given that the point has $x = 2$.

So, The required point is $\left(2 , f \left(2\right)\right) = \left(2 , {2}^{3} - {e}^{2}\right) = \left(2 , 8 - {e}^{2}\right)$.............(i)

Now, We need to calculate the slope of the graph on that point.

So, The equation of slope for $f \left(x\right)$ is:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} - {e}^{x}\right) = \frac{d}{\mathrm{dx}} {x}^{3} - \frac{d}{\mathrm{dx}} {e}^{x} = 3 {x}^{2} - {e}^{x}$

Now. we will just substitute $x = 2$ in $f ' \left(x\right)$ to get the slope.

So, The slope of $f \left(x\right)$ at $x = 2$ is:

$f ' \left(2\right) = 3 \times {2}^{2} - {e}^{2} = 12 - {e}^{2}$.

Now, we know, Condition of Perpendicularity of two lines is ${m}_{1} {m}_{2} = - 1$ where ${m}_{1}$ and ${m}_{2}$ are the slopes of the lines respectively.

So, The slope of our normal line is $- \frac{1}{12 - {e}^{2}}$.

Now, Let's find the equation.

We know the Slope-Intercept Form of a linear equation is:

$y = m x + c$ [Where, $m$ is the slope and $c$ is the y-intercept.]

In this equation, let's substitute our slope.

So, The equation becomes,

$y = - \frac{1}{12 - {e}^{2}} \times x + c \Rightarrow y = - \frac{x}{12 - {e}^{2}} + c$.........................(ii)

And, This line must pass through our desired point $\left(2 , 8 - {e}^{2}\right)$.

So, the equation must be satisfied with those values.

So, Now substituting the values of $x$ and $y$, solve for $c$.

So, We get,

$8 - {e}^{2} = - \frac{2}{12 - {e}^{2}} + c \Rightarrow c = \frac{2}{12 - {e}^{2}} - 8 + {e}^{2}$.

Now, We will put $c$ in eq (ii) to get the answer.

So, We get,

$y = \frac{2 - x}{12 - {e}^{2}} - 8 + {e}^{2}$ ........................(iii)

Hope this helps.