Question

At what point is the graph of y=1/2x^2 is the tangent line para llel to the line 2x-4y=3?

Answer 1

first, m(slope) = y' = 1/2 * 2x = x

see that the slope line 2x-4y=3 is
m = -a/b
(here a = 2, b = -4, and c = 3)
so,
m = -2/-4 = 1/2,

back to first, get
x = 1/2

then put x = 1/2 on the graph of y = 1/2 x^2, get
y = 1/2 * (1/2)^2 = 1/8

therefore, the point is (x,y) = (1/2, 1/8)