What effect will introducing a constant have on an exponential function?

I got this answer from Kalahira (quality assurance)

( user) I just wanted to give credit.

y = e^x is an exponential function that increases gradually from (-âž, 0) until it reaches the point (0,1). Thereafter, it very sharply increases on (0,âž).

If you introduce a constant A > 1,

y = Ae^x will increase gradually from (-âž, 0) until it reaches the point (0,A).

Thereafter, it will rise faster than the base function y = e^x

Example. y = 5e^x

If your constant A is less than 1 but still positive: 0 < A < 1,

y = Ae^x will increase gradually from (-âž, 0) until it reaches the point (0,A).

Thereafter, it will have a slower rise on (0,âž) than the base function y = e^x.

Example. y = (1/5) e^x

If your constant A = -1,

y = Ae^x will look like y = Ae^x when A=1 except that the graph will be flipped over the x-axis.

Instead of increasing gradually on (-âž, 0), it will decrease gradually on (-âž, 0)

until it passes through the point (0, -1) and then veer downward on (0,âž)

Example. y = -e^x

If your constant A is between -1 and 0: -1 < A < 0,

y = Ae^x will be flipped over the x-axis and decrease gradually from(-âž, 0)

until it reaches the point (0,A).

Thereafter, it will have a slower decrease on (0,âž)

Example. y = (-1/5) e^x

If your constant A is less than 0: A < -1,

y = Ae^x will be reflected (flipped) over the x-axis and decrease gradually from (-âž, 0)

until it reaches the point (0,A). Thereafter, it will sharply veer downward.

Example. y = -5 e^x

y = e^x + A will translate the base function y = e^x by A units. If A > 0, it will essentially shift the graph upward by A. If A < 0, it will shift the graph downward by A.

Example. y = e^x + 5

Example. y = e^x - 5

Step-by-step explanation:

answer: can you me with homework?

step-by-step explanation:

you can't solve x because it is 2 linear equation and you only have one equation