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

Question

through the P(1,4)

Answer 1

For any exponential function, the horizontal asymptote is always at y=c (where c is the constant in your formula). So, c=-2.

Notice that -2 is 4 units below 2 (the y-intercept). When we set x=0, we find:

f(0)=b-2=2
b=4. The distance between the horizontal asymptote and y-intercept is the coefficient b. (You make it negative if the y-intercept is below the asymptote.)

So far, we have f(x) = 4a^x-2.

We want it to pass through (1,4), so we set
4=4a^1-2
6=4a
a=6/4=3/2=1.5

That is, f(x)=4(1.5)^x-2.

When x is a very large negative number, the exponential term decays to nothing, leaving us with just the horizontal asymptote, y=-2. Also, f(0)=4-2=2, and f(1)=4(1.5)-2=6-2=4. Quit 'n Eat Dinner.

Answer 2

f(0) = ba^0 + c = b + c = 2 f(a million) = ba^a million + c = ab + c = 4 so ab - b = 2 via fact the horizontal asymptote is y = -2, all of us be conscious of 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

asymptote y=-2 → c=-2.
intercept → f(0) = b+c = 2
Through (1,4) → f(1) = ba+c = 4.

Three equations, three variables. Solve for a,b,c.