Verified answer

a) Differentiate f(x) by power rule to get...

f(x) = 8x^(-1) + 2x

f'(x) = -8x^(-2) + 2

If f'(x) = 0, then...

0 = -8/x² + 2

8/x² = 2

8 = 2x²

x² = 8/2

x = ±√4

x = {2} [Since x > 0.]

By first derivative sign test, we can show that f'(x) changes from negative to positive at x = 2. Since this happens, we can show (2,8) is the turning point [decreasing to increasing]

b) Vertical asymptote occurs where the denominator is 0, so it's x = 0.

To find the horizontal asymptote, combine both terms and use the ∞ limit...

f(x) = 8/x + 2x * x/x

= (8/x) + 2x²/x

= (8 + 2x²)/x

Since the degree for the numerator is greater than that for the denominator, we can show that the limits contain ∞. Really, asymptote never contains ∞, so there is no horizontal asymptote.

You can see the graph here... http://www.wolframalpha.com/input/?i=8%2Fx+%2B+2x+...

I hope this helps!

a) f'(x) = -8/x^2 + 2, so f'(x) = 0 at x = 2. Therefore, (2,8) is a minimum.

b) The line x = 0 is a vertical asymptote.