Question

identify the graph of y = e^x - 2.

Explanation:

Step1: Analyze the parent function \( y = e^x \)

The parent function \( y = e^x \) has a y - intercept at \( (0,1) \) (when \( x = 0 \), \( y=e^0 = 1 \)) and it is an exponential growth function, increasing as \( x \) increases and approaching \( y = 0 \) as \( x ightarrow-\infty \).

Step2: Analyze the transformation for \( y=e^x - 2 \)

The function \( y = e^x-2 \) is a vertical shift of the parent function \( y = e^x \) down by 2 units. So, we need to find the y - intercept of \( y = e^x-2 \). When \( x = 0 \), \( y=e^0 - 2=1 - 2=- 1 \). Also, as \( x ightarrow-\infty \), \( e^x ightarrow0 \), so \( y = e^x - 2 ightarrow0 - 2=-2 \). And as \( x ightarrow\infty \), \( e^x ightarrow\infty \), so \( y = e^x - 2 ightarrow\infty \).

Now let's analyze the given graphs:
- The first graph has a y - intercept at \( (0,0) \) (since it touches the y - axis at \( y = 0 \))? Wait, no, looking at the first graph, when \( x = 0 \), the y - value seems to be 0? Wait, no, the first graph: when \( x = 0 \), the curve is at \( y = 0 \)? Wait, no, the parent function \( y = e^x \) has y - intercept 1, and \( y=e^x - 2 \) has y - intercept - 1. Wait, maybe I misread. Wait, let's check the second graph: when \( x = 0 \), \( y=-1 \)? Wait, the second graph: when \( x = 0 \), the curve passes through \( (0, - 1) \)? Wait, no, the second graph's curve at \( x = 0 \) is at \( y=-1 \)? Wait, the second graph: let's see the y - axis. The second graph has a curve that, as \( x ightarrow-\infty \), approaches \( y=-2 \) (since it's a vertical shift down by 2 of \( y = e^x \), which approaches \( y = 0 \) as \( x ightarrow-\infty \), so \( y=e^x - 2 \) approaches \( y=-2 \) as \( x ightarrow-\infty \)), and at \( x = 0 \), \( y=e^0-2=-1 \), and it's an increasing function (since \( y = e^x \) is increasing and vertical shift doesn't change the monotonicity).

The first graph: when \( x = 0 \), the y - value is 0 (since the curve touches the y - axis at \( y = 0 \)), which would correspond to \( y=e^x - 1 \), not \( y=e^x - 2 \). The third graph is a decreasing function (it's a reflection or a different exponential function, like \( y = e^{-x} \) or something), so it's not an exponential growth. The second graph: as \( x ightarrow-\infty \), it approaches \( y=-2 \), at \( x = 0 \), \( y=-1 \), and it's increasing, which matches the properties of \( y = e^x-2 \).

Wait, maybe I made a mistake earlier. Let's re - evaluate:

For \( y = e^x-2 \):

- Domain: all real numbers.
- Range: \( (-2,\infty) \) (since \( e^x>0 \), so \( e^x - 2>-2 \))
- y - intercept: when \( x = 0 \), \( y=1 - 2=-1 \)
- As \( x ightarrow-\infty \), \( y ightarrow - 2 \)
- As \( x ightarrow\infty \), \( y ightarrow\infty \)

Now check the graphs:

- First graph: y - intercept at \( y = 0 \) (when \( x = 0 \), the curve is at \( y = 0 \)), range seems to be \( (0,\infty) \) (since it's above \( y = 0 \)), so it's \( y = e^x \) (since \( y = e^x \) has range \( (0,\infty) \) and y - intercept 1? Wait, no, the first graph's y - intercept is 0? Maybe the first graph is \( y = e^x-1 \).

- Second graph: as \( x ightarrow-\infty \), it approaches \( y=-2 \), at \( x = 0 \), \( y=-1 \), and it's increasing. The range is \( (-2,\infty) \), which matches \( y = e^x-2 \).

- Third graph: it's a decreasing function (as \( x \) increases, \( y \) decreases), which would be a function like \( y = e^{-x}-2 \) or something, but our function is \( y = e^x-2 \) which is increasing.

So the correct graph should be the second one (the middle graph) because it has the correct y - intercept (\( y=-1 \) when \( x = 0 \)), approaches \( y=-2 \) as \( x ightarrow-\infty \), and is an increasing exponential function.

Answer:

The middle graph (the second graph among the three given graphs) is the graph of \( y = e^x-2 \).