please help me with this question: i havent got the answer yet?

Question

find the equation of the tangent to the curve y = 2x - x^2 that is perpendicular to the line x + 4y - 8 = 0

Answer 1

The slopes of lines that are perpendicular are negative reciprocals of each other.

Find the slope of the x + 4y - 8 = 0 by putting it into y = mx + b form

x + 4y = 8
4y = -x + 8
y = (-1/4)x + 8

The slope is -1/4

The slope of a line that is perpendicular is 4 (negative reciprocal)

The slopes of the lines tangent to the given curve is given by the first derivative.

y = 2x - x^2
y' = 2 - 2x

Set this equal to 4

2 - 2x = 4
-2x = 2
x = -1

So, at x = -1, the slope of the tangent line to the give curve is 4.
Find y at x = -1
y = 2(-1) - (-1)^2
y = -2 - 1
y = -3

Find the equation of the line with slope 4 and goes through the point (-1,-3)

y = mx + b
-3 = 4(-1) + b
-3 = -4 + b
1 = b

y = 4x + 1 <===== there you go!

Answer 2

a line perpendicular to x + 4y - 8 = 0 (which has a slope of -1/4) must have a slope of 4. slope of the tangent to y = 2x - x² is the derivative, y' = 2 - 2x. so
2 - 2x = 4
-2x = 2
x = -1
when x = -1, y = 2(-1) - (-1)² = -2 - 1 = -3
so we want the line through (-1,-3) that has a slope of 4, and in point-slope form that's y + 3 = 4(x + 1), which can be written as y = 4x + 1.

Answer 3

sinc eit is perpendicular to x+4y-8, the slope is 4 (negative reciprocal). you weren't given a point?? if oyu have a point its easy but im not sure without