Find the equation of a line normal to a given function, given its slope (no points.)

Question

I have no idea how to solve this without points.

Find the equation of the line, normal to $y=(2-x)^4$, which has slope equal to $-\dfrac{1}{32}$

If it helps, the solution is $y=(-1/32)x+(129/8)$

But I truly have no idea how to solve this.

Answer 1

$y=(-x+2)^4$

$y'= 4(-x+2)^3(-1)$ [By Product Rule]

$y'=-4(2-x)^3$

$y' = 32$ due to being the negative reciprocal of the slope of the normal line

$32=-4(2-x)^3$

$x=4$

$y=16$

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

$y-16 = -(1/32)(x-4)$

$y=-(1/32)x+(129/8)$

Answer 2

The slope of the normal is given as $- \dfrac{1}{32} $, therefore, the slope of the tangent line is $32$

Now, the derivative is $ y' = 4 (x - 2)^3 $

Equate this expression to $32$ you get

$(x - 2)^3 = 8 $

Hence $x = 4$, and from this, $y = (2)^4 = 16 $

Now we have the point $(4, 16)$ and the slope of the normal line,

so its equation is

$ y - 16 = - \frac{1}{32} (x - 4 ) $

Which becomes

$ y = - \dfrac{1}{32} x + 16 + 1/8 = - \dfrac{1}{32} x + \dfrac{129}{8} $