Question

f(x)=x³+6x²-36x-216
this appears to be a polynomial of degree 2.
yes no
this function appears to have a positive leading coefficient.
yes no
this function appears to be f(x).
yes no

this appears to be a polynomial of degree 2.
yes no
this function appears to have a positive leading coefficient.
yes no
this function appears to be f(x).
yes no

this appears to be a polynomial of degree 2.
yes no
this function appears to have a positive leading coefficient.
yes no
this function appears to be f(x).
yes no

Explanation:

First, analyze the given function \( f(x) = x^3 + 6x^2 - 36x - 216 \). This is a cubic function (degree 3), so any graph claiming it's a degree 2 (quadratic) polynomial is incorrect.

For the leading coefficient: the leading term is \( x^3 \) with coefficient 1 (positive). So as \( x \to \infty \), \( f(x) \to \infty \), and as \( x \to -\infty \), \( f(x) \to -\infty \) (since odd degree with positive leading coefficient).

Now, check each graph:

1. First graph: Let's see the end behavior. Wait, but the function is cubic (degree 3), so it should have two turning points (since degree \( n \) has at most \( n - 1 \) turning points). A degree 2 (quadratic) has 1 turning point. So the first question "This appears to be a polynomial of degree 2" – answer No. "Positive leading coefficient" – let's see end behavior. For cubic with positive leading coefficient, as \( x \to \infty \), up; \( x \to -\infty \), down. Wait, maybe I misread. Wait the function is \( x^3 + 6x^2 - 36x - 216 \). Let's factor it: \( x^2(x + 6) - 36(x + 6) = (x^2 - 36)(x + 6) = (x - 6)(x + 6)(x + 6) \). So roots at \( x = 6 \), \( x = -6 \) (double root). So the graph touches the x-axis at \( x = -6 \) (double root) and crosses at \( x = 6 \). So the graph should have a double root at \( x = -6 \) (so touches there) and crosses at \( x = 6 \).

Now, let's analyze each graph:

First graph: "This appears to be a polynomial of degree 2" – No (since it's cubic, degree 3). "Positive leading coefficient" – let's see end behavior. For cubic with positive leading coefficient, as \( x \to \infty \), \( f(x) \to \infty \); as \( x \to -\infty \), \( f(x) \to -\infty \). So the first graph's end behavior: if it's a cubic, but the question is about degree 2. Wait, the first sub-question for each graph:

First graph:

- "This appears to be a polynomial of degree 2": No (since \( f(x) \) is degree 3).
- "This function appears to have a positive leading coefficient": Let's see end behavior. If the graph goes up on the right (as \( x \to \infty \)) and down on the left (as \( x \to -\infty \)), that's positive leading coefficient for cubic. So yes? Wait, but the first question is degree 2 – No. Then "This function appears to be \( f(x) \)": Let's check roots. \( f(x) \) has a double root at \( x = -6 \) and single at \( x = 6 \). So the graph should touch at \( x = -6 \) and cross at \( x = 6 \).

Second graph: Degree 2? No (since \( f(x) \) is degree 3). Positive leading coefficient? If it's a parabola (degree 2) opening down, then leading coefficient negative. So "positive leading coefficient" – No. "This function appears to be \( f(x) \)" – No, since \( f(x) \) is cubic, not quadratic.

Third graph: Degree 2? No. Positive leading coefficient? For cubic, positive leading coefficient means right end up, left end down. Let's see the third graph's end behavior: right end up, left end down? Wait, the third graph: as \( x \to \infty \), up; \( x \to -\infty \), down? Wait, maybe. Then "This function appears to be \( f(x) \)" – let's check roots. \( f(x) \) has a double root at \( x = -6 \) (so touches x-axis there) and single at \( x = 6 \). So the graph should have a touch at \( x = -6 \) (so the graph touches the x-axis, not crosses) and crosses at \( x = 6 \). So the third graph: does it touch at a root? Let's see.

Wait, maybe I made a mistake. Let's re-express:

Function: \( f(x) = (x + 6)^2(x - 6) \). So it's a cubic with a double root at \( x = -6 \) (so the graph touches the x-axis at \( x = -6 \), turning around there) and a single root at \( x = 6 \) (crosses the x-axis there). So the graph should have a local maximum or minimum at \( x = -6 \) (since double root) and another turning point. So the graph will touch the x-axis at \( x = -6 \), go up, then down, then cross at \( x = 6 \) and go up (since leading coefficient positive). Wait, no: let's compute the derivative: \( f'(x) = 3x^2 + 12x - 36 = 3(x^2 + 4x - 12) = 3(x + 6)(x - 2) \). So critical points at \( x = -6 \) and \( x = 2 \). So at \( x = -6 \), it's a local maximum or minimum? Let's check the second derivative: \( f''(x) = 6x + 12 \). At \( x = -6 \), \( f''(-6) = 6(-6) + 12 = -24 < 0 \), so local maximum at \( x = -6 \). At \( x = 2 \), \( f''(2) = 6(2) + 12 = 24 > 0 \), so local minimum at \( x = 2 \). So the graph: at \( x = -6 \), local max (touches x-axis? Wait, \( f(-6) = 0 \), so the local max is at \( ( -6, 0 ) \), so the graph touches the x-axis there (since it's a double root), then decreases to a local min at \( x = 2 \), then increases, crossing the x-axis at \( x = 6 \) (since \( f(6) = 0 \)). Wait, no: \( f(6) = 6^3 + 6*6^2 - 36*6 - 216 = 216 + 216 - 216 - 216 = 0 \), correct. \( f(2) = 8 + 24 - 72 - 216 = 32 - 288 = -256 \), so local min at \( (2, -256) \). So the graph: starts from bottom left (as \( x \to -\infty \), \( f(x) \to -\infty \)), goes up to \( x = -6 \) (local max, \( y = 0 \)), then down to \( x = 2 \) (local min, \( y = -256 \)), then up to top right (as \( x \to \infty \), \( f(x) \to \infty \)). Wait, that's the end behavior: left end down, right end up (since positive leading coefficient, odd degree). So the graph should have left end down, right end up, a local max at \( x = -6 \) (on x-axis), local min at \( x = 2 \) (below x-axis), then up through \( x = 6 \) (on x-axis).

Now, check the three graphs:

First graph: Let's see end behavior. If left end is up, right end is down – that's negative leading coefficient (cubic with negative leading coefficient: \( x \to \infty \), \( f(x) \to -\infty \); \( x \to -\infty \), \( f(x) \to \infty \)). So first graph: end behavior left up, right down – negative leading coefficient. So "positive leading coefficient" – No. "This appears to be \( f(x) \)" – No, because end behavior is wrong (f(x) has positive leading coefficient, so left down, right up).

Second graph: It's a parabola (degree 2), so has 1 turning point. But f(x) is degree 3, has 2 turning points. So "This appears to be a polynomial of degree 2" – No. "Positive leading coefficient" – if it's a parabola opening down, leading coefficient negative – No. "This appears to be \( f(x) \)" – No, since it's quadratic.

Third graph: End behavior: left end down, right end up (positive leading coefficient, cubic). Number of turning points: 2 (matches cubic). Roots: touches at \( x = -6 \) (local max on x-axis), crosses at \( x = 6 \) (after local min). So this should be the graph of \( f(x) \).

So let's answer each sub-question for the third graph (assuming the third graph is the one with left end down, right end up, two turning points, touching at a root and crossing at another):

1. "This appears to be a polynomial of degree 2": No (since f(x) is degree 3).
2. "This function appears to have a positive leading coefficient": Yes (end behavior left down, right up, positive leading coefficient for cubic).
3. "This function appears to be \( f(x) \)": Yes (matches end behavior, number of turning points, and root behavior: touch at \( x = -6 \), cross at \( x = 6 \)).

But maybe I mixed up the graphs. Wait the problem has three graphs, each with three questions:

For each graph:

- Q1: "This appears to be a polynomial of degree 2" – answer No (all, since f(x) is degree 3).
- Q2: "This function appears to have a positive leading coefficient" – check end behavior. For f(x), positive leading coefficient (odd degree) means left end down, right end up. So which graph has that? The third graph (assuming the third one has left down, right up).
- Q3: "This function appears to be \( f(x) \)" – check if it has a double root at \( x = -6 \) (touching x-axis) and single root at \( x = 6 \) (crossing), with two turning points. The third graph should satisfy this.

But maybe the user is asking for each graph's answers. Let's assume the three graphs are:

Graph 1:

- "This appears to be a polynomial of degree 2": No
- "This function appears to have a positive leading coefficient": Let's see end behavior. If left up, right down – No. If left down, right up – Yes. Wait maybe I made a mistake earlier. Wait the function is \( x^3 + 6x^2 - 36x - 216 \), leading coefficient 1 (positive), degree 3 (odd). So end behavior: as \( x \to \infty \), \( f(x) \to \infty \); as \( x \to -\infty \), \( f(x) \to -\infty \) (left down, right up). So a graph with left down, right up has positive leading coefficient (cubic). A graph with left up, right down has negative leading coefficient (cubic: \( -x^3 + ... \)).

So Graph 1: If it's left up, right down – negative leading coefficient – so "positive leading coefficient" – No.

Graph 2: It's a parabola (degree 2), so "This appears to be a polynomial of degree 2" – No (since f(x) is degree 3). "Positive leading coefficient" – if it's opening down, leading coefficient negative – No.

Graph 3: Left down, right up (positive leading coefficient, cubic), two turning points, touches at \( x = -6 \), crosses at \( x = 6 \) – so "This appears to be \( f(x) \)" – Yes.

So for the third graph:

- "This appears to be a polynomial of degree 2": No
- "This function appears to have a positive leading coefficient": Yes
- "This function appears to be \( f(x) \)": Yes

But maybe the user is asking for each graph's answers. Let's proceed step by step.

Assuming the three graphs are labeled as Top, Middle, Bottom:

Top Graph:

1. "This appears to be a polynomial of degree 2": No (f(x) is degree 3)
2. "This function appears to have a positive leading coefficient": No (end behavior left up, right down – negative leading coefficient)
3. "This function appears to be \( f(x) \)": No (end behavior and roots don't match)

Middle Graph:

1. "This appears to be a polynomial of degree 2": No (f(x) is degree 3, this is quadratic)
2. "This function appears to have a positive leading coefficient": No (parabola opening down – negative leading coefficient)
3. "This function appears to be \( f(x) \)": No (it's quadratic)

Bottom Graph:

1. "This appears to be a polynomial of degree 2": No (f(x) is degree 3)
2. "This function appears to have a positive leading coefficient": Yes (end behavior left down, right up – positive leading coefficient, cubic)
3. "This function appears to be \( f(x) \)": Yes (matches end behavior, number of turning points, roots)

So the answers for the bottom graph (third graph) would be:

- "This appears to be a polynomial of degree 2": No
- "This function appears to have a positive leading coefficient": Yes
- "This function appears to be \( f(x) \)": Yes

But maybe the user is asking for each graph's set of questions. Let's assume the three graphs are:

First graph (top):

1. Is it degree 2? No
2. Positive leading coefficient? No (end behavior left up, right down)
3. Is it f(x)? No

Second graph (middle):

1. Is it degree 2? No
2. Positive leading coefficient? No (parabola opening down)
3. Is it f(x)? No

Third graph (bottom):

1. Is it degree 2? No
2. Positive leading coefficient? Yes
3. Is it f(x)? Yes

So for each graph, the answers are:

Top Graph:

- "This appears to be a polynomial of degree 2": No
- "This function appears to have a positive leading coefficient": No
- "This function appears to be \( f(x) \)": No

Middle Graph:

- "This appears to be a polynomial of degree 2": No
- "This function appears to have a positive leading coefficient": No
- "This function appears to be \( f(x) \)": No

Bottom Graph:

- "This appears to be a polynomial of degree 2": No
- "This function appears to have a positive leading coefficient": Yes
- "This function appears to be \( f(x) \)": Yes

But maybe the user is asking for the third graph's answers. Let's confirm the function's graph:

\( f(x) = (x + 6)^2(x - 6) \). So:

- Degree 3, so not degree 2: No
- Leading coefficient positive (1), so end behavior left down, right up: Yes (if the graph has that)
- Roots at \( x = -6 \) (double, touches x-axis) and \( x = 6 \) (single, crosses x-axis): Yes (if the graph touches at -6 and crosses at 6)

So the answers for the third graph (assuming it's the one with left down, right up, two turning points, touching at a root and crossing at another) are:

1. No
2. Yes
3. Yes

Answer:

For the third graph (assuming it's the bottom one):

• "This appears to be a polynomial of degree 2": No
• "This function appears to have a positive leading coefficient": Yes
• "This function appears to be \( f(x) \)": Yes

(If the graphs are labeled differently, adjust based on end behavior and roots, but the key is the function is cubic with positive leading coefficient, degree 3, roots at \( x = -6 \) (double) and \( x = 6 \) (single).)