I need to write 2 equations that represent the same exponential function with a y-intercept of 5 and an asymptote at y=3. I got y=2^(x+1) + 3 but I don't know how to find the second equation. Can someone please explain this to me. Thanks.
I don't know at what level you are working at, but any exponential of the form y = p^x can be written as
y = a(e^kx) where a and k are constants.
So let your y=2^(x+1) + 3 be
y = a(e^kx) + 3
let's use 2 points from your first equation, say (0,5) and (1,7)
in the new one:
for (0,5) ,
5 = a(e^0) + 3
a = 2
for (1,7)
7 = 2(e^k) + 3
2 = e^k
k = ln2
so a second equation would be
y = 2(e^[(ln2)x)] + 3
check it by using (2,11) from the first equation and trying it in our new one.
LS = 11
RS = 2(e^2ln2) + 3
= 2(e^1.386294361) + 3
= 2(4) + 3
= 11
= RS
Thank you so much for your detailed reply. My problem is that we haven't done logs or "e" yet. Is there any other way arrive at an answer without knowing these? I really appreciate your help.
To find the second equation representing the same exponential function, we need to use the given information about the y-intercept and the asymptote.
The general form of an exponential function is y = a * b^x + c, where "a" represents the vertical stretch or compression, "b" represents the base, and "c" represents the vertical shift.
Given that the y-intercept is 5, we know that the point (0, 5) lies on the graph of the exponential function. Plugging these values into the equation, we have:
5 = a * b^0 + c
5 = a * 1 + c
5 = a + c --------(Equation 1)
On the other hand, given that the asymptote is at y = 3, it means that the graph of the exponential function approaches but never reaches y = 3. Therefore, as x approaches infinity, y should approach 3. This leads us to conclude that the vertical shift, or "c" in the general form, should be 3.
Now we can substitute the value of c = 3 into Equation 1:
5 = a + 3
a = 5 - 3
a = 2
Now we have the values for "a" and "c". Plugging them into the general form of the exponential function, we can write the second equation:
y = 2 * b^x + 3
So, the second equation representing the same exponential function with a y-intercept of 5 and an asymptote at y = 3 is y = 2 * b^x + 3.
Note that the value of the base, "b," can vary as long as it is a positive number greater than 1 for an increasing exponential function.