Question

Give the equation of the exponential function whose graph is shown. HINTS: (1) There are no stretches or shrinks. (2) Look at just the graph and asymptote to determine any reflections and/or vertical shifts. (3) Use the labeled points to determine the base of the exponential function and any horizontal shifts. Coordinates, (-3,(-1/e)+1), (-2,0), (-1,-e+1), (0,-e^2+1) and on the graph there is a horizontal dotted line going through 1 on the y axis (0,1)

Answer 1

So the hints help us get a few things. An exponential graph with coordinates points containing e's. So definitely \(e^{x}\) is a starting point. The last hint says we have a horizontal asymptote at y = 1, which automatically indicates a vertical shift of 1, giving us the function \(e^{x} + 1\). The last little bit is easy to get if we just use one of the coordinate points.

What I'm going to do is assume we have the function \(ae^{x} + 1\), and I'm going to use the point (-2,0). This means when I plug in -2 for x, I should get an answer of 0. So now I'll solve for a using this:
\(ae^{-2} + 1 = 0\)
\(ae^{-2} = -1\)
\(a=-e^{2}\)
Using this value of a, I have:
\(-e^{2}\cdot e^{x} + 1 = -e^{x+2} + 1\) using exponent rules. And we can check the other points you have with this function
For x = -3

\(-e^{-3+2}+1 = \frac{-1}{e} + 1\)
We already did x = -2

For x = -1

\(-e^{-1+2}+1 = -e+1\)

For x= 0

\(-e^{0+2}+1 = -e^{2} + 1\)

So all the conditions are satisfied and the function is \(-e^{x+2}+1\)

Answer 2

Great! Thank you so much, that's very detailed :)