Question 159204
I'll do the first one to get you started


a)


a)g(x) = e^x + 3



Graph: vertical shift of 3 units up. Correct.



Equation for horizontal asymptote:  When "x" is a very large negative number, then {{{e^x}}} becomes really small (and eventually approaches zero). 


So what's left over is the term "3". So the horizontal asymptote is {{{y=3}}}



y-intercept: 


{{{g(x)=e^x+3}}} Start with the given function



{{{g(0)=e^0+3}}} Plug in {{{x=0}}}



{{{g(0)=1+3}}} Raise "e" to the zeroth power to get 1



{{{g(0)=4}}} Add



So when {{{x=0}}} then {{{y=4}}} (g(x) and y are interchangeable) 



So the y-intercept is (0,4)



Here's a graph to verify the answer.


{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,exp(x)+3)

)}}} Graph of {{{g(x)=e^x+3}}}



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b)


Use the techniques used above to solve this problem.