Question

What is the equation of the normal line of `f(x)=x^3-49x^2+7x` at `x=7`?

Answer 1

`y=1/532x-2009.013`

Explanation:

The normal line at a point is the line perpendicular to the tangent line at that point. When we solve problems of this type, we find the slope of the tangent line using the derivative, use that to find the slope of the normal line, and use a point from the function to find the normal line equation.

Step 1: Slope of the Tangent Line
All we do here is take the derivative of the function and evaluate it at `x=7`:
`y' = 3x^2-98x+7`
`y'(7) = 3(7)^2-98(7)+7`
`y'(7) = -532`

That means the slope of the tangent line at `x=7` is -532.

Step 2: Slope of the Normal Line
The slope of the normal line is simply the opposite inverse of the slope of the tangent line (because these two are perpendicular). So we just flip -532 and make it positive to get `1/532` as the slope of the normal line.

Final Step: Finding the Equation
Normal line equations are of the form `y=mx+b`, where `y` and `x` are points on the line, `m` is the slope, and `b` is the `y`-intercept. We have the slope, `m`, which is what we found in step two: `1/532`. The points `x` and `y` can be easily found by substituting `x=7` into the equation and solving for `y`:
`y=(7)^3-49(7)^2+7(7)`
`y = -2009`

Now we can use all this information to find `b`, the `y`-intercept:
`y=mx+b`
`-2009 = (1/532)(7)+b`
`-2009 = 7/532+b`
`-2009-7/532 = b`
We can approximate this to -2009.013, or if we really wanted to, we could approximate it too -2009.

The equation of the normal line is thus `y=1/532x-2009.013`.