find the exponential function of the form f (x) = b a^x + c with y - intercept 2, horizontal asymptote y = -2,

Answer 1

Ok, let's do it :

f(x) = b*a^x + c

y - intercept 2, means, that for x = 0, then y = 2

f(0) = 2 = b + c

horizontal asymptote y = -2, means :

Lim ( b*a^x + c) = -2

x-----+/-infinity

a^x = e^(a*lnx)

Lim ( b*e^a*lnx + c)

if x --- - infinity, then :

b*a(1/infinity) + c = -2

b*0 + c = -2

c = -2

and b = 4

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

Hope that helps

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 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.

Answer 3

y = f(x) = ba^x + c

The y intercept is 2.

f(0) = 2 = bx^0 + c = b*1 + c = b + c
2 = b + c
c = 2 - b
a ≠ 0

The horizontal asymptote is y = -2. This is the value f(x) approaches as x→∞.

f(∞) = -2 = ba^∞ + c = ba^∞ + (2 - b)

The value of the function will be infinite unless |a| ≤ 1.

If |a| ≤ 1, then a^∞ = 0.

And we are left with:

-2 = 2 - b
b = 4
c = 2 - b = 2 - 4 = -2

So the function is

y = f(x) = 4a^x - 2
where | a | < 0 and a ≠ 0.