find exponential function form f(x)=ba^x+c with y- intercept 2, horizontal asymptote y=-2 passes point P(1,4)?

Answer 1

Your goal here is to solve for a, b, and c. We can easily do this with the facts that we're given.

y-intercept is 2: This means that when x=0, f(0) = 2

so f(0) = b * a^(0) + c = 2
b + c = 2 (since a^0 = 1)

Horizontal asymptote is y = -2. Remember that for exponential functions, the horizontal asymptote is always the horizontal shift, which is the constant next to the term containing the expontential. Thus, in our case, c is the horizontal shift, so

c = -2

Combining this with b+c=2 (which is what we solved for earlier), substituting c=-2 in there yields:

b+(-2) = 2, so b=4.

b=4 and c = -2, so if we plug it back into f(x), we get

f(x) = 4*a^x - 2

Passes through the point (1,4) means we plug in 1 for x and 4 for f(x), to get:

4 = 4*a^(1) - 2, or
4 = 4a - 2
6 = 4a
3/2 = a

Now we have a, b, and c. The final answer is then:

f(x) = 4*(3/2)^x - 2

Answer 2

f(0) = ba^0 + c = b + c = 2 f(a million) = ba^a million + c = ab + c = 4 so ab - b = 2 because of the fact the horizontal asymptote is y = -2, all of us comprehend ba^x, which has y=0 as horizontal asymptote, has been translated down 2, so c = -2. then b - 2 = 2, b = 4, and 4a - 4 = 2 4a = 6 a = a million.5 so f(x) = 4(a million.5^x) - 2, answer d.