Question 835546
I'm guessing that the vertical asymptote occurs at {{{x=-2}}} which means the function doesn't exist there so {{{x+2}}} must be in the denominator.
The horizontal asymptote occurs at {{{y=5}}}. 
SO the function looks something like,
{{{f(x)=5((x+a)/(x+2))}}}
when {{{x=0}}}, {{{y=7}}}
{{{f(x)=5((0+a)/(0+2))=7}}}
{{{(5a)/2=7}}}
{{{a=14/5}}}
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{{{f(x)=5((x+14/5)/(x+2))}}}
{{{highlight(f(x)=((5x+14)/(x+2)))}}}
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{{{ drawing(300,300,-10,10,-10,10,grid(1),line(-2,20,-2,-20),graph( 300, 300, -10, 10, -10, 10, 5((x+14/5)/(x+2)),5)) }}}