Question

Find an exponential function with a horizontal asymptote y equals=2 whose graph contains the points ​(0,3​) and ​(1,6​).

Answer 1

To find an exponential function with a horizontal asymptote y = 2 that passes through the points (0, 3) and (1, 6), the function is y = 3 * 2^x + 2.

Step-by-step explanation:

To find an exponential function with a horizontal asymptote y = 2 that passes through the points (0, 3) and (1, 6), we can use the general form of an exponential function, y = a * b^x, where a is the initial value and b is the growth or decay factor.

Using the given points, we can form a system of equations:

3 = a * b^0 = a

6 = a * b^1 = ab

From the first equation, we know that a = 3. Substituting this value into the second equation gives us 6 = 3b, which simplifies to b = 2.

Therefore, the exponential function that satisfies the given conditions is y = 3 * 2^x + 2.

Answer 2

`\bf y=a(b)^x\qquad \qquad (\stackrel{x}{0}~,~\stackrel{y}{3})\implies 3=a(b)^0\implies 3=a \\\\\\ therefore\qquad y=3(b)^x\\\\ -------------------------------\\\\ (\stackrel{x}{1}~,~\stackrel{y}{6})\implies 6=3(b)^1\implies \cfrac{6}{3}=b^1\implies 2=b \\\\\\ therefore\qquad y=3(2)^x`

now, exponential functions have a horizontal asymptote at the x-axis, namely when y = 0, however, if you just move this one with a vertical translation of 2, then the horizontal asymptote will be at 2 instead. y = 3(2)ˣ + 2