Mark M. answered 08/12/15
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Retired math prof. Very extensive Precalculus tutoring experience.
Start with a function of the form y = a(2)x, where a is a real number to be determined. Any function of this type has horizontal asymptote y = 0 (the x-axis).
To get a horizontal asymptote of y = 4, translate the graph of
y = a(2)x four units upward. The equation of the translated graph is y = a(2)x + 4.
We want this graph to have y-intercept (0, 2). So, 2 = a(2)0 + 4.
Simplifying, we get 2 = a + 4 Thus, a = -2.
We have: y = -2(2)x + 4
Note: 2x > 0 for all x So, -2(2)x < 0 for all x
Therefore, -2(2)x + 4 < 4 for all x
So, the range of y = -2(2)x + 4 is y < 4
y = -2(2)x + 4 is a function with the desired properties.
We can use properties of exponents to rewrite the function as
y = -(2)x+1 + 4