Question

what is the average rate of change where x ∈ 0, 1?
what is the average rate of change where x ∈ -1, 1?
what is the equation of the asymptote? (do not use spaces)

Explanation:

Step 1: Identify the function (exponential, \( y = 2^x \) or similar)

The graph passes through \( (0, 2) \) (wait, no—wait, at \( x = 0 \), the y - value: looking at the grid, when \( x = 0 \), the point is at \( y = 2 \)? Wait, no, maybe \( y = 2^x \)? Wait, no, let's check points. At \( x = 0 \), the y - coordinate: from the graph, when \( x = 0 \), the curve is at \( y = 2 \)? Wait, no, maybe \( y = 2^x \)? Wait, no, let's re - examine. Wait, the horizontal asymptote is \( y = 0 \) (since the left - hand side approaches \( y = 0 \)). Let's find the y - values at the endpoints of the intervals.

Step 2: Average rate of change for \( x \in [0, 1] \)

The formula for average rate of change of a function \( f(x) \) over \( [a,b] \) is \( \frac{f(b)-f(a)}{b - a} \).
Assume the function is \( f(x)=2^x \). Then \( f(0)=2^0 = 1 \)? Wait, no, from the graph, at \( x = 0 \), the y - value is 2? Wait, maybe the function is \( y = 2^x+1 \)? Wait, no, the horizontal asymptote is \( y = 0 \), so as \( x\to-\infty \), \( y\to0 \). At \( x = 0 \), let's say the point is \( (0, 2) \), and at \( x = 1 \), the point is \( (1, 4) \) (since the curve is exponential and increasing).
So for \( [0,1] \): \( a = 0 \), \( b = 1 \), \( f(0)=2 \), \( f(1)=4 \). Then average rate of change \(=\frac{4 - 2}{1-0}=\frac{2}{1}=2 \).

Step 3: Average rate of change for \( x \in [-1, 1] \)

First, find \( f(-1) \). If the function is \( y = 2^x+1 \)? No, wait, if the horizontal asymptote is \( y = 0 \), and \( f(0)=2 \), then maybe \( f(x)=2^x + 0 \)? Wait, no, at \( x=-1 \), if \( f(x)=2^x \), then \( f(-1)=\frac{1}{2}=0.5 \), \( f(0)=1 \)? Wait, I think I made a mistake earlier. Let's re - correct. Let's assume the function is \( f(x)=2^x \). Then:

• At \( x = 0 \), \( f(0)=2^0 = 1 \)? But the graph at \( x = 0 \) seems to be at \( y = 2 \). Wait, maybe the function is \( f(x)=2^{x + 1} \). Then \( f(0)=2^{0 + 1}=2 \), \( f(1)=2^{1+1}=4 \), \( f(-1)=2^{-1 + 1}=2^{0}=1 \).

Now, for \( x\in[-1,1] \): \( a=-1 \), \( b = 1 \), \( f(-1)=1 \), \( f(1)=4 \). The average rate of change \(=\frac{4 - 1}{1-(-1)}=\frac{3}{2}=1.5 \).

Step 4: Equation of the asymptote

The horizontal asymptote: as \( x\to-\infty \), the graph approaches \( y = 0 \), so the equation of the horizontal asymptote is \( y = 0 \).

Answer:

For \( x \in [0, 1] \): \( 2 \)
For \( x \in [-1, 1] \): \( 1.5 \)
Equation of asymptote: \( y = 0 \)