document.write( "Question 1209884: What is the horizontal asymptote as x approaches positive infinity of the graph of
\n" ); document.write( "y = \sqrt{4x^2 + 5x} - \sqrt{4x^2}?
\n" ); document.write( "The horizontal asymptote is in the form y = mx + k.
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Algebra.Com's Answer #850957 by greenestamps(13198)\"\" \"About 
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\n" ); document.write( "!!!! 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.

\n" ); document.write( "Ignoring that (or allowing the slope m to be 0)....

\n" ); document.write( "\"y=sqrt%284x%5E2%2B5x%29-sqrt%284x%5E2%29\"

\n" ); document.write( "Rationalize the numerator:

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\n" ); document.write( "\"y=%28%284x%5E2%2B5x%29-%284x%5E2%29%29%2F%28sqrt%284x%5E2%2B5x%29%2Bsqrt%284x%5E2%29%29%29\"

\n" ); document.write( "\"y=%285x%29%2F%28sqrt%284x%5E2%2B5x%29%2Bsqrt%284x%5E2%29%29%29\"

\n" ); document.write( "\"y=%285x%29%2F%282x%2Asqrt%281%2B5%2F4x%29%2B2x%29\"

\n" ); document.write( "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

\n" ); document.write( "\"y=%285x%29%2F%282x%2B2x%29=%285x%29%2F%284x%29=5%2F4\"

\n" ); document.write( "ANSWER: \"y=5%2F4\"

\n" ); document.write( "(or \"y=0x%2B5%2F4\"....)

\n" ); document.write( "A graph....

\n" ); document.write( "\"graph%28400%2C300%2C-10%2C50%2C-1%2C2%2Csqrt%284x%5E2%2B5x%29-sqrt%284x%5E2%29%29\"

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