Question

What is the equation of the normal line of `f(x)= x^8 + 6 e^x` at `x=0`?

Answer 1

`y-6=-1/6(x-0)` in the slope point form. `y=-1/6x + 6` in slope intercept form. The detailed explanation is given below.

Explanation:

We need to find the equation of the normal line.

To find the equation of the line we need two things.
1. Slope `m`.
2. Point `(x_1,y_1)`
Then the equation can be formed by using the slope point form

`y-y_1 = m(x-x_1)`

We are to find the equation of normal at `x=0` let us use that to find `f(0)` this would give our points `(x_1,y_1)`
`f(x)=x^8+6e^x`
`f(0) = 0^8 +6e^0`
`f(0) = 0+6`
`f(0)=6`

Our `(x_1,y_1) = (0,6)`

Now to find the slope of required equation. For this, we need to find the slope of the tangent. The slope of the tangent is found by finding `f'(x)`

`f'(x)=8x^7+6e^x`
Slope of tangent at `x=0`
`f'(0) = 8(0)^7+6e^0`
`f'(0)=6`

The slope of the normal is the negative reciprocal of slope of the tangent.

Slope of normal `m = -1/6`

Equation of the normal line at `x=0` is

`y-6=-1/6(x-0)` in the slope point form.

`y=-1/6x + 6` in slope intercept form.