Question

What is the equation of the line normal to `f(x)=x^2+8x - 12` at `x=3`?

Answer 1

`x+32y=675`

Explanation:

Given:
`color(white)("XXX")f(x)=x^2+8x-12`

At `x=3`
`color(white)("XXX")f(3)=3^2+8(3)-12=21`

The slope of `f(x)=x^2+8x-12` for an arbitrary `x` is given by the derivative:
`color(white)("XXX")m_x=f'(x)=2x+8`

At `x=3` the slope is
`color(white)("XXX")m_3=f'(3)=2(3)+8=32`

If `f(x)` has a slope of `m`
then a line perpendicular to it (which the normal is) has a slope of `(-1/m)`

Therefore the normal to `f(x)` at `x=3`
passes through the point `(3,21)` and a has a slope of `(-1/32)`

Using the slope-point form for a line
`color(white)("XXX")y-21=-1/32(x-3)`

Converting to standard form:
`color(white)("XXX")32y-672=-x+3`

`color(white)("XXX")x+32y=675`

graph{(x^2+8x-12-y)(x+32y-675)=0 [-66, 65.6, -29, 36.85]}