Question

At what point on the graph of y=(1/2)x^2-(3/2) is the tangent line parallel to the line 4x-8y=5?

A. (1/2,-3/8)
B. (1/2,-11/8)
C. (2,3/8)
D. (2,1/2)
E. (-1/2,-11/8)

Answer 1

y = (1/2)x^2 - (3/2) -> this is a parabola
differentiate:
dy/dx = (1/2)*2*x - 0
dy/dx = x

The line 4x - 8y = 5 has a slope of,
-8y = -4x + 5
y = (1/2)x - 5/8
slope = 1/2
If this line and the tangent line of the parabola are parallel, they have the same slope:
dy/dx = x
1/2 = x
Substituting this to the parabola equation to get y,
y = (1/2)x^2 - (3/2)
y = (1/2)(1/2)^2 - 3/2
y = 1/8 - 12/8
y = -11/8

Thus point is at (1/2, -11/8).

Hope this helps :3