Question

1. describe the end behavior of the graph.

image of the graph of y = 3^x
a. as x → +∞, f(x) → +∞ and x → -∞, f(x) → -∞
b. as x → +∞, f(x) → +∞ and x → -∞, f(x) → 0
c. as x → +∞, f(x) → -∞ and x → -∞, f(x) → -∞
d. as x → +∞, f(x) → 0 and x → -∞, f(x) → -∞

1. using the graph below, identify the interval(s) on which f(x) is positive and the interval(s) on which f(x) is negative.

image of the graph of f(x) = 2^(x+1) - 3
a. f(x) is positive on (0.5, ∞) and f(x) is negative on (-∞, 0.5)
b. f(x) is positive on (-∞, 0) and f(x) is negative on (0.5, ∞)
c. f(x) is positive on (-1, ∞) and f(x) is negative on (-1, ∞)
d. f(x) is positive on (-∞, -1) and f(x) is negative on (-∞, -1)

Explanation:

Question 3

Step1: Analyze right end behavior

For the exponential function \( y = 3^x \), as \( x \to +\infty \), \( 3^x \) grows without bound, so \( f(x) \to +\infty \).

Step2: Analyze left end behavior

As \( x \to -\infty \), \( 3^x=\frac{1}{3^{|x|}} \), and as \( |x| \) becomes large, \( \frac{1}{3^{|x|}} \to 0 \) (since the denominator grows to infinity). So we check the options: option b has \( x\to+\infty, f(x)\to+\infty \) and \( x\to-\infty, f(x)\to 0 \), which matches.

Step1: Find the x - intercept

The function is \( f(x)=2^{x + 1}-3 \). To find where \( f(x)=0 \), solve \( 2^{x+1}-3 = 0\Rightarrow2^{x + 1}=3\Rightarrow x + 1=\log_2{3}\Rightarrow x=\log_2{3}-1\approx1.58 - 1 = 0.58\approx0.5 \) (approximate value).

Step2: Analyze the graph's position

The graph of an exponential function \( y = 2^{x+1}-3 \) is an exponential growth (since the base \( 2>1 \)) shifted down by 3 and left by 1. For \( x>0.5 \) (approximate x - intercept), the graph is above the x - axis (so \( f(x)>0 \)), and for \( x < 0.5 \), the graph is below the x - axis (so \( f(x)<0 \)). Option a states that \( f(x) \) is positive on \( (0.5,\infty) \) and negative on \( (-\infty,0.5) \), which matches.

Answer:

b. As \( x \to +\infty, f(x) \to +\infty \) and \( x \to -\infty, f(x) \to 0 \)

Question 4