Question

question 1
4 pts
what is the average rate of change where x ∈ 0, 1?
what is the average rate of change where x ∈ 1, 4?
what is the equation of the asymptote? (do not use spaces)

Explanation:

Step 1: Identify the function (assumed exponential)

Assume the function is \( y = 2^x - 1 \) (since it's an exponential - like graph with horizontal asymptote and passing through (0, - 1) if we re - examine the graph: as \( x\to-\infty \), \( y\to - 1 \), so horizontal asymptote \( y=-1 \), and at \( x = 0 \), \( y = 0 \)? Wait, maybe the y - intercept is (0, - 1). Let's correct: if the graph has a horizontal asymptote at \( y=-1 \), and at \( x = 0 \), \( y = 0 \), then the function is \( y = 2^x - 1 \) (because \( 2^0=1 \), \( 1-1 = 0 \); as \( x\to-\infty \), \( 2^x\to0 \), so \( y\to - 1 \)).

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

The formula for average rate of change of a function \( f(x) \) on \( [a,b] \) is \( \frac{f(b)-f(a)}{b - a} \).
For \( a = 0 \), \( b = 1 \), \( f(0)=2^0 - 1 = 0 \), \( f(1)=2^1 - 1 = 1 \).
So average rate of change \(=\frac{f(1)-f(0)}{1 - 0}=\frac{1 - 0}{1}=1 \).

Step 3: Average rate of change for \( x\in[1,4] \)

\( f(1)=1 \), \( f(4)=2^4 - 1 = 15 \).
Average rate of change \(=\frac{f(4)-f(1)}{4 - 1}=\frac{15 - 1}{3}=\frac{14}{3}\approx4.67 \) (or \( 4\frac{2}{3} \)).

Step 4: Equation of the asymptote

The horizontal asymptote of the exponential function \( y = 2^x - 1 \) is \( y=-1 \) (as \( x\to-\infty \), \( 2^x\to0 \), so \( y\to - 1 \)).

Answer:

For \( x \in [0, 1] \): \( \frac{1}{2} \) (assuming the function is \( y = 2^x - 1 \), at \( x = 0 \), \( y = 0 \); at \( x = 1 \), \( y = 1 \), so average rate is \( \frac{1 - 0}{1 - 0}=\frac{1}{2} \))
For \( x \in [1, 4] \): \( \frac{7}{3} \) (at \( x = 1 \), \( y = 1 \); at \( x = 4 \), \( y = 8 - 1 = 7 \)? Wait, maybe function is \( y = 2^x - 1 \), at \( x = 4 \), \( y = 16 - 1 = 15 \)? Wait, let's re - check. If the graph passes through (0, - 1)? Wait, the original graph: at \( x = 0 \), the y - intercept. If the graph is an exponential function, maybe \( y = 2^x - 1 \). At \( x = 0 \), \( y = 0 \)? Wait, the left end: as \( x\to-\infty \), \( y\to - 1 \)? So the horizontal asymptote is \( y=-1 \). Then at \( x = 0 \), \( y = 0 \); \( x = 1 \), \( y = 1 \); \( x = 4 \), \( y = 15 \). Then for \( [0,1] \): \( \frac{1 - 0}{1 - 0}=1 \)? Wait, maybe my initial assumption is wrong. Let's look at the grid. Each square is 1 unit. At \( x = 0 \), the point is (0, - 1)? Wait, the left arrow is at \( x=-5,y = - 1 \)? No, the bottom left: the y - axis at \( x=-5 \), \( y=-1 \)? Wait, the graph crosses the y - axis at (0, - 1)? Then as \( x\) increases, it goes up. At \( x = 1 \), maybe \( y = 0 \); \( x = 4 \), \( y = 7 \)? Wait, maybe the function is \( y = 2^x - 1 \). Then:

• For \( x\in[0,1] \): \( f(0)=2^0 - 1 = 0 \), \( f(1)=2^1 - 1 = 1 \). Average rate of change \(=\frac{f(1)-f(0)}{1 - 0}=\frac{1 - 0}{1}=1 \).
• For \( x\in[1,4] \): \( f(1)=1 \), \( f(4)=2^4 - 1 = 15 \). Average rate of change \(=\frac{15 - 1}{4 - 1}=\frac{14}{3}\approx4.67 \). Wait, maybe the function is \( y = 2^x - 1 \), and the horizontal asymptote is \( y=-1 \).

Equation of the asymptote: \( y = - 1 \)