SOLUTION: What is the horizontal asymptote as x approaches positive infinity of the graph of y = \sqrt{4x^2 + 5x} - \sqrt{4x^2}? The horizontal asymptote is in the form y = mx + k.

Algebra ->  Graphs -> SOLUTION: What is the horizontal asymptote as x approaches positive infinity of the graph of y = \sqrt{4x^2 + 5x} - \sqrt{4x^2}? The horizontal asymptote is in the form y = mx + k.       Log On


   

Question 1209884: What is the horizontal asymptote as x approaches positive infinity of the graph of
y = \sqrt{4x^2 + 5x} - \sqrt{4x^2}?
The horizontal asymptote is in the form y = mx + k.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


!!!! The equation of a horizontal asymptote is not in the form y = mx + k, unless you are letting m be 0. The equation of a horizontal asymptote is of the form y = k.

Ignoring that (or allowing the slope m to be 0)....

y=sqrt%284x%5E2%2B5x%29-sqrt%284x%5E2%29

Rationalize the numerator:



y=%28%284x%5E2%2B5x%29-%284x%5E2%29%29%2F%28sqrt%284x%5E2%2B5x%29%2Bsqrt%284x%5E2%29%29%29

y=%285x%29%2F%28sqrt%284x%5E2%2B5x%29%2Bsqrt%284x%5E2%29%29%29

y=%285x%29%2F%282x%2Asqrt%281%2B5%2F4x%29%2B2x%29

As x goes to positive infinity, 5%2F4x goes to 0 so sqrt%281%2B5%2F4x%29 goes to sqrt%281%29 = 1, and the expression approaches

y=%285x%29%2F%282x%2B2x%29=%285x%29%2F%284x%29=5%2F4

ANSWER: y=5%2F4

(or y=0x%2B5%2F4....)

A graph....

graph%28400%2C300%2C-10%2C50%2C-1%2C2%2Csqrt%284x%5E2%2B5x%29-sqrt%284x%5E2%29%29