Question 1209884
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!!!! 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.<br>
Ignoring that (or allowing the slope m to be 0)....<br>
{{{y=sqrt(4x^2+5x)-sqrt(4x^2)}}}<br>
Rationalize the numerator:<br>
{{{y=(sqrt(4x^2+5x)-sqrt(4x^2))*((sqrt(4x^2+5x)+sqrt(4x^2))/(sqrt(4x^2+5x)+sqrt(4x^2)))}}}<br>
{{{y=((4x^2+5x)-(4x^2))/(sqrt(4x^2+5x)+sqrt(4x^2)))}}}<br>
{{{y=(5x)/(sqrt(4x^2+5x)+sqrt(4x^2)))}}}<br>
{{{y=(5x)/(2x*sqrt(1+5/4x)+2x)}}}<br>
As x goes to positive infinity, {{{5/4x}}} goes to 0 so {{{sqrt(1+5/4x)}}} goes to {{{sqrt(1)}}} = 1, and the expression approaches<br>
{{{y=(5x)/(2x+2x)=(5x)/(4x)=5/4}}}<br>
ANSWER: {{{y=5/4}}}<br>
(or {{{y=0x+5/4}}}....)<br>
A graph....<br>
{{{graph(400,300,-10,50,-1,2,sqrt(4x^2+5x)-sqrt(4x^2))}}}<br>