-->graph the function.\n24. $f(x)=2e^{x}$\n25. $f(x)=\frac{1}{2}e^{x - 2}$\n26. $f(x)=-e^{2x}+2$

Explanation:

Step1: Identify key - points for (f(x)=2e^{x})

When (x = 0), (f(0)=2e^{0}=2\times1 = 2). As (x\to-\infty), (f(x)\to0), and as (x\to+\infty), (f(x)\to+\infty). Plot the point ((0,2)) and draw a curve that approaches the (x) - axis as (x\to-\infty) and increases rapidly as (x\to+\infty).

Step2: Identify key - points for (f(x)=\frac{1}{2}e^{x - 2})

First, consider the horizontal shift. The graph of (y = e^{x}) is shifted 2 units to the right. When (x = 2), (f(2)=\frac{1}{2}e^{2 - 2}=\frac{1}{2}). As (x\to-\infty), (f(x)\to0), and as (x\to+\infty), (f(x)\to+\infty). Plot the point ((2,\frac{1}{2})) and draw a curve with the appropriate end - behavior.

Step3: Identify key - points for (f(x)=-e^{2x}+2)

When (x = 0), (f(0)=-e^{0}+2=-1 + 2=1). As (x\to-\infty), (f(x)\to2) (since (e^{2x}\to0) as (x\to-\infty)), and as (x\to+\infty), (f(x)\to-\infty) (since (e^{2x}\to+\infty) as (x\to+\infty)). Plot the point ((0,1)) and draw a curve with the correct end - behavior.

To graph these functions accurately on the given grids:

• For (y = 2e^{x}), start at the (y) - intercept ((0,2)) and draw an increasing exponential curve that gets closer and closer to the (x) - axis as (x) goes to negative infinity.
• For (y=\frac{1}{2}e^{x - 2}), start at the point ((2,\frac{1}{2})) and draw an increasing exponential curve with the same general shape as (y = e^{x}) but shifted 2 units to the right.
• For (y=-e^{2x}+2), start at the (y) - intercept ((0,1)), draw a decreasing curve that approaches (y = 2) as (x) goes to negative infinity and goes down towards negative infinity as (x) goes to positive infinity.

The actual graphing is done on the provided grids by hand - plotting the key points and drawing smooth curves through them. Since this is a graphing problem, we can't give a single numerical answer. The task is to create the visual representation of the functions on the given coordinate grids.