Question

use the graph to determine open interval on which the function is increasing, if any. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the function is increasing on the interval(s) . (type your answer in interval notation. use a comma to separate answers as needed.) b. there is no interval on which the function is increasing.

Explanation:

Step1: Identify the vertex

The graph is a parabola opening upwards. The vertex is at \( x = 0 \) (from the graph, the lowest point is at \( (0, -1) \) approximately, but the axis of symmetry is \( x = 0 \)).

Step2: Determine increasing interval

For a parabola opening upwards, the function increases to the right of the vertex. The vertex is at \( x = 0 \), so the function is increasing on the open interval \( (0, \infty) \)? Wait, no, looking at the graph, the vertex seems to be at \( x = 0 \) (the minimum point is at \( x = 0 \)). Wait, the graph: let's check the x-axis. The grid: from -5 to 5. The parabola has its minimum at \( x = 0 \) (since it's symmetric around x=0). So to the right of x=0, as x increases, y increases. So the open interval where the function is increasing is \( (0, \infty) \)? Wait, no, the graph's x-axis: the minimum is at x=0 (the point (0, -1) maybe). Wait, looking at the graph, the left side comes from the top left, decreases to (0, -1), then increases to the top right. So the function is increasing on \( (0, \infty) \)? Wait, but the x-axis goes to 5, but the interval is open. Wait, no, the standard parabola \( y = x^2 - 1 \) would have vertex at (0, -1), and increasing on \( (0, \infty) \). So the open interval is \( (0, \infty) \)? Wait, but the options: option A says "the interval(s)". Wait, maybe I misread the graph. Wait, the graph: the minimum is at x=0 (the point (0, -1)), so the function is increasing on \( (0, \infty) \)? Wait, no, let's check the x-values. Wait, the graph's x-axis: from -5 to 5. The parabola: when x increases from 0 to 5, y increases. So the open interval is \( (0, \infty) \)? But maybe the graph is \( y = (x)^2 - 1 \), so the increasing interval is \( (0, \infty) \). Wait, but the problem says "open interval", so we use (0, ∞). But maybe the graph is actually with vertex at x=0, so the increasing interval is (0, ∞). Wait, but let's check again. The graph: the left side (x < 0) is decreasing, right side (x > 0) is increasing. So the open interval where the function is increasing is \( (0, \infty) \). Wait, but the options: option A is to fill in the interval. So the correct interval is \( (0, \infty) \)? Wait, no, maybe the vertex is at x=0, so the increasing interval is (0, ∞). So the answer is option A, with interval (0, ∞). Wait, but maybe the graph is actually with vertex at x=0, so the increasing interval is (0, ∞). So the open interval is \( (0, \infty) \).

Answer:

A. The function is increasing on the interval(s) \((0, \infty)\)