Question

1. aubrey is examining the graph of a function, shown below; aubrey writes the following statements about the graph. select the three statements that are correct. a. this function is continuous. b. this function has domain {x|x ≥ -9} c. this function is increasing. d. this function has a local minimum. e. this function has an x -intercept at (4, 0) f. this function has a y -intercept at (0, -4)

Explanation:

Step1: Analyze Option A

A continuous function has no breaks, gaps, or holes. The graph shown is a smooth curve without any discontinuities, so the function is continuous. Thus, A is correct.

Step2: Analyze Option B

The domain of a function is all possible x - values. Looking at the graph, the left - most point seems to start around \(x=- 9\) (from the visual of the graph), so the domain is \(\{x|x\geq - 9\}\). So, B is correct.

Step3: Analyze Option C

A function is increasing if as \(x\) increases, \(y\) always increases. The graph has a part where it decreases (the left - hand curve) and then increases (the right - hand curve), so it is not always increasing. Thus, C is incorrect.

Step4: Analyze Option D

A local minimum is a point where the function changes from decreasing to increasing. The right - hand part of the graph has a "valley" which is a local minimum. So, D is correct.

Step5: Analyze Option E

An x - intercept is a point where \(y = 0\). Looking at the graph, the x - intercept on the right seems to be at \(x = 4\) (since when \(x = 4\), \(y=0\)), so the x - intercept is \((4,0)\). Thus, E is correct? Wait, no, let's re - check. Wait, the graph: the left curve crosses the x - axis, and the right curve also crosses. Wait, maybe I made a mistake. Wait, the original graph: let's look again. Wait, the right - hand curve: when \(x = 4\)? Wait, maybe the x - intercept is at \(x = 4\)? Wait, no, let's check the options again. Wait, the options: E says x - intercept at \((4,0)\). Let's see the graph: the right - hand curve crosses the x - axis at \(x = 4\) (since the grid: from the graph, the x - axis crossing on the right is at \(x = 4\)). Wait, but also, let's check the y - intercept. The y - intercept is where \(x = 0\). From the graph, when \(x = 0\), the y - value is around - 8? Wait, no, the option F (wait, the options: A, B, C, D, E, F? Wait, the user's image: the options are A: continuous, B: domain \(x\geq - 9\), C: increasing, D: local minimum, E: x - intercept \((4,0)\), F: y - intercept \((0,-4)\). Wait, let's re - evaluate:

Wait, the graph: the function is a polynomial - like graph. Let's re - check each option:

- A: The graph is a single smooth curve, so it's continuous. Correct.
- B: The left - most point of the graph (the start of the domain) is at \(x=-9\) (from the visual, as the graph starts at \(x = - 9\) approximately), so domain \(x\geq - 9\). Correct.
- C: The graph decreases, then increases, so not always increasing. Incorrect.
- D: The right - hand "valley" is a local minimum (where the function changes from decreasing to increasing). Correct.
- E: x - intercept is where \(y = 0\). The right - hand x - intercept is at \(x = 4\) (since on the x - axis, at \(x = 4\), \(y = 0\)), so \((4,0)\) is an x - intercept. Correct? Wait, but let's check the y - intercept. The y - intercept is at \(x = 0\). From the graph, when \(x = 0\), the y - value is not - 4. So F is incorrect.

Wait, the problem says "Select the THREE statements that are correct". Wait, maybe my initial analysis was wrong. Wait, let's start over:

1. Option A: The graph is a continuous curve (no breaks), so A is correct.
2. Option B: The domain: the graph starts at \(x=-9\) (the left - most point), so all \(x\geq - 9\). Correct.
3. Option C: The graph has a decreasing interval (left part) and an increasing interval (right part), so it's not increasing everywhere. Incorrect.
4. Option D: The graph has a local minimum (the lowest point on the right - hand curve, where the function changes from decreasing to increasing). Correct.
5. Option E: x - intercept: the right - hand x - intercept is at \(x = 4\) (since when \(x = 4\), \(y = 0\)). Correct? Wait, but also, let's check the y - intercept. The y - intercept is at \(x = 0\). From the graph, when \(x = 0\), the y - value is - 8 (approx), so F (y - intercept \((0,-4)\)) is incorrect.

Wait, but the problem says "Select the THREE statements that are correct". So A, B, D? Or A, D, E? Wait, maybe I made a mistake with B. Wait, the domain of a function is all real numbers for a polynomial, but this graph seems to have a restricted domain? Wait, no, maybe the graph is a piecewise function? Wait, the original graph: the left - hand curve and the right - hand curve. Wait, maybe the domain is all real numbers? Wait, I think I made a mistake with B. Let's re - consider:

Wait, the graph: it's a continuous curve, so domain is all real numbers? No, the left - hand curve starts at \(x=-10\)? Wait, the grid: the x - axis goes from - 10 to 10. The left - hand curve starts at \(x=-10\) (the arrow goes down at \(x=-10\)). Wait, maybe the domain is all real numbers. Then B is incorrect. Oh, I see my mistake. The graph: the left - hand curve starts at \(x=-10\) (the arrow is at \(x=-10\)), so the domain is all real numbers, so B is incorrect.

Let's start over:

- A: The graph is a single unbroken curve, so it's continuous. Correct.
- B: Domain: the graph extends to the left as \(x\) approaches \(-\infty\)? No, the left - hand curve has an arrow going down at \(x=-10\), but maybe it's a polynomial of degree 3 or 4? Wait, no, the graph has two "humps": a maximum on the left and a minimum on the right. So it's a cubic or quartic? Wait, no, the graph is continuous everywhere, so domain is all real numbers. So B is incorrect.
- C: The graph is not increasing (it decreases, then increases), so C is incorrect.
- D: The right - hand "valley" is a local minimum (the function decreases to that point then increases), so D is correct.
- E: The x - intercept on the right: from the graph, the right - hand curve crosses the x - axis at \(x = 4\) (since the grid: each square is 1 unit? So when \(x = 4\), \(y = 0\)), so \((4,0)\) is an x - intercept. Correct.
- F: The y - intercept is where \(x = 0\). From the graph, when \(x = 0\), the y - value is - 8 (not - 4), so F is incorrect.

Wait, but the problem says "Select the THREE statements that are correct". So A, D, E? Wait, no, let's check again:

Wait, the graph:

- A: Continuous: yes, because there are no breaks. Correct.
- D: Local minimum: yes, the right - hand low point. Correct.
- E: x - intercept at \((4,0)\): yes, the graph crosses the x - axis at \(x = 4\), \(y = 0\). Correct.

Wait, but also, is there a local maximum on the left? Yes, but the option D is local minimum. Also, A is correct. So A, D, E? Wait, no, let's check the original options again. The options are:

A. This function is continuous.

B. This function has domain \(\{x|x\geq - 9\}\)

C. This function is increasing.

D. This function has a local minimum.

E. This function has an x - intercept at \((4,0)\)

F. This function has a y - intercept at \((0,-4)\)

So the correct statements are A, D, E? Wait, no, let's check the y - intercept again. When \(x = 0\), the y - value: looking at the graph, the point at \(x = 0\) is at \(y=-8\) (since the grid: from \(y = - 10\) to \(y = 10\), and the minimum on the right is at \(y=-8\)? Wait, no, the graph at \(x = 0\) is between \(y=-5\) and \(y=-10\). Wait, maybe the y - intercept is \((0,-8)\), so F is incorrect.

So the correct three statements are A, D, E? Wait, no, let's check A: continuous, correct. D: local minimum, correct. E: x - intercept at \((4,0)\), correct. Also, is B correct? If the graph starts at \(x=-9\), then B is correct. Maybe the left - most point is at \(x=-9\) (the graph starts at \(x=-9\) instead of \(x=-10\)). So if the domain is \(x\geq - 9\), then B is correct.

Given the initial marking (B is marked), maybe the intended domain is \(x\geq - 9\). So let's assume that the left - most point is at \(x=-9\), so B is correct. Then the correct three are A, B, D? Or A, D, E?

Wait, the problem says "Select the THREE statements that are correct". Let's use the process of elimination:

- C is wrong (not increasing).
- F is wrong (y - intercept not \((0,-4)\)).

So we have A, B, D, E left. We need three.

- A: correct (continuous).
- D: correct (local minimum).
- E: correct (x - intercept \((4,0)\)).

Or A, B, D:

- A: correct.
- B: correct (domain \(x\geq - 9\)).
- D: correct (local minimum).

But let's check the x - intercept: the right - hand curve crosses the x - axis at \(x = 4\), so E is correct. The y - intercept: at \(x = 0\), the y - value is - 8 (not - 4), so F is wrong.

So the correct three are A, D, E? Wait, no, maybe the x - intercept is at \(x = 4\), so E is correct. The local minimum (D) is correct. Continuous (A) is correct. So A, D, E.

But the initial marking has B marked, so maybe the domain is \(x\geq - 9\). Given the confusion, but following the visual:

The correct statements are A (continuous), D (local minimum), and E (x - intercept \((4,0)\)) or A, B, D.

Wait, let's look at the graph again:

- The graph is continuous (A: correct).
- It has a local minimum (D: correct).
- It has an x - intercept at \((4,0)\) (E: correct).

So the three correct statements are A, D, E.

Answer:

A. This function is continuous, D. This function has a local minimum, E. This function has an \(x\) - intercept at \((4,0)\)