a) f ' (2) = 20(2) - 6(2)^2 - 16 = 40 - 24 - 16 = 0
= the slope of the x axis......so.....parallel
The equation of this tangent line is y = 0 (x -2) + 3 .... y = 3 (1)
b) The slope of the tangent line at x = 3 is
20(3) - 6(3)^2 - 16 = 60 - 54 - 16 = -10
So....the slope of the normal line is (1/10)
The equation of this normal line is
y = (1/10)(x - 3) -1
y = (1/10)x - 3/10 - 1
y = (1/10)x - 13/10 sub (1) into this to find the x coordinate of R
3 = (1/10)x - 13/10
3 + 13/10 = x /10
43/10 = x /10
43 = x
So.....R = (43, 3)
See the graph here to confirm this :
https://www.desmos.com/calculator/0ipyvcpc5y
P.S. - If you have had Integral Calculus....you might see how I determined f(x)
