Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passe

Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4).


a. f(x)=-2(2^x)


b. f(x)=2(2^x) -2


c. f(x)=2(1.5^x)-2


d.f(x)=4(1.5^x)-2

Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passe
y = ba^x + c





If the horizontal asymptote is y = -2, then c = -2.





y = ba^x - 2





Since 2 is the y-intercept, then


2 = ba^0 - 2 which simplifies to


2 = b - 2


Add 2 to each side.


4 = b





Now we have y = 4(a^x) - 2


Now let's use the last piece of information.


(1,4) is a point on the graph, so


4 = 4(a^1) -2


4 = 4a - 2


Add 2 to each side


6 = 4a


Divide both sides by 4


6/4 = a


Reduce.


a = 3/2





y = 4(3/2)^x - 2

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