Question

the polynomial function f(x) is graphed below. fill in the form below regarding the features of this graph.
answer attempt 1 out of 2
the degree of f(x) is and the leading coefficient is . there are
different real zeros and relative minimums.

Explanation:

Step1: Determine the degree (even/odd)

The ends of the polynomial graph both go down (as \( x \to \pm\infty \), \( f(x) \to -\infty \)). For a polynomial, if the leading coefficient is negative and the degree is even, the ends both go down. So the degree is even (let's check the number of turns: the graph has 3 turning points. The maximum number of turning points of a polynomial is \( n - 1 \), so \( n - 1 = 3 \implies n = 4 \)? Wait, no: number of turning points is at most \( n - 1 \). Wait, the graph has 3 turning points? Wait, looking at the graph: first a peak, then a valley, then a peak, then down. Wait, no: the graph crosses the x-axis on the left, then has a peak, a valley (at y-axis), a peak, then crosses the x-axis on the right. So the number of turning points: from left, it comes up from bottom, crosses x-axis, goes to a peak, then down to a valley (at y-axis, touching? No, it's a valley, not touching), then up to a peak, then down. So turning points: 3? Wait, no: the derivative's roots. Wait, the end behavior: both ends down, so leading term is \( -ax^n \), \( n \) even. Now, number of real zeros: it crosses the x-axis twice (left and right), so 2 real zeros? Wait, no: the graph crosses the x-axis on the left (one zero) and on the right (another zero), so 2 distinct real zeros? Wait, no, maybe I miscounted. Wait, the graph: starts from bottom left, crosses x-axis (zero 1), goes up to a local max, down to a local min (at y-axis, but does it touch x-axis? No, the local min is above or on x-axis? Wait, the graph at the local min (the middle one) is on the y-axis, touching x-axis? Wait, the graph shows that the local min is at (0,0)? Wait, no, the graph crosses the x-axis on the left, then has a peak, then a valley (which is on the x-axis? Wait, the problem's graph: let's re-examine. The graph: left end down, crosses x-axis (zero 1), goes up to a local maximum, down to a local minimum (which is on the y-axis, maybe at (0,0)?), then up to a local maximum, then down, crossing x-axis again (zero 2). Wait, no, if the local minimum is on the x-axis, then that's a repeated zero. But the question says "different real zeros". So the graph crosses the x-axis twice (two distinct real zeros) or three? Wait, no, the left crossing, then the middle is a touch (repeated zero), then right crossing? But the graph as drawn: the left part crosses x-axis, then goes up, down to a valley (which is on x-axis, so that's a zero with multiplicity 2), then up, then down, crossing x-axis again. So total real zeros: 3? Wait, no, the number of times it crosses or touches the x-axis. But the question is "different real zeros", so distinct ones. So if it crosses at two distinct points (left and right) and touches at the middle (same as one of them? No, middle is at y-axis, so x=0. So if the local min is at x=0, y=0, then that's a zero at x=0, and another at left, another at right? Wait, no, the graph: left end down, crosses x-axis (x = a < 0), goes up to local max, down to local min at (0,0) (so x=0 is a zero), then up to local max, then down, crossing x-axis at x = b > 0. So three real zeros? But the local min at x=0 is on the x-axis, so that's a zero. So different real zeros: 3? Wait, no, the question says "different real zeros", so distinct. So x = a, x=0, x=b: three distinct? But maybe the local min is not on x-axis. Wait, the original graph: the user's graph: let's see the arrows. Left end down, crosses x-axis (zero 1), up to local max, down to local min (which is above x-axis? No, the graph shows that the local min is at y=0? Wait, the graph's middle part: the local min is on the x-axis, so that's a zero. So three real zeros? But the question's answer boxes: "There are [ ] different real zeros". Wait, maybe I made a mistake. Let's go back to end behavior: both ends down, so degree is even (since even degree with negative leading coefficient: ends both down). Number of turning points: the graph has 3 turning points (local max, local min, local max), so degree is at least 4 (since turning points = n - 1, so n = 4? Wait, 3 turning points: n - 1 = 3 ⇒ n = 4. So degree 4 (even), leading coefficient negative (since ends down). Now, real zeros: how many times does it cross x-axis? Let's see: left side: crosses x-axis (1), then local max, local min (which is on x-axis? So that's a touch, so multiplicity 2), then local max, then crosses x-axis (2). Wait, no, if the local min is on x-axis, then that's a zero with multiplicity 2, so total real zeros: 2 distinct (the left and right crossings, and the middle is a repeated zero of the same as one? No, middle is at x=0, so if left is x = -c, middle is x=0, right is x = c, then three distinct zeros? But the graph as drawn: maybe the local min is above x-axis, so it only crosses x-axis twice (left and right), so two distinct real zeros. Then relative minimums: how many local minima? The graph has two local minima: one at the middle (the lower one) and maybe? Wait, no: the graph has a local maximum, then a local minimum (the first min), then a local maximum, then a local minimum (the second min)? Wait, no, the graph: from left, comes up (crosses x-axis), goes to a local maximum (first max), then down to a local minimum (first min), then up to a local maximum (second max), then down to a local minimum (second min)? Wait, no, the end behavior is both down, so the degree is even, number of turning points: 3 (max, min, max, min? No, 3 turning points: max, min, max. Wait, no, the graph has 3 turning points: first a local max, then a local min, then a local max. Wait, that's 3 turning points, so degree 4 (since n - 1 = 3 ⇒ n = 4). So local minima: how many? The graph has one local min (the middle one) or two? Wait, no: a local minimum is a point where the function changes from decreasing to increasing. So after the first local max (increasing to decreasing), then decreasing to increasing (local min), then increasing to decreasing (local max), then decreasing (since end is down). Wait, no, the graph: left end down, so as x increases from -∞, f(x) increases (crosses x-axis), reaches local max (turning point 1: max), then decreases to local min (turning point 2: min), then increases to local max (turning point 3: max), then decreases to -∞. So there is 1 local min? Wait, no, two local minima? Wait, no, the first min is after the first max, then the second min is after the second max? No, the second max is the last peak, then it goes down, so no min after that. Wait, I'm confused. Let's use the end behavior and turning points.

Degree: even (since both ends down), leading coefficient negative.

Number of real zeros: the graph crosses the x-axis twice (left and right), so 2 different real zeros (assuming the middle min is above x-axis, so it doesn't cross there).

Relative minima: the graph has one local minimum (the middle one) or two? Wait, the graph has a local maximum, then a local minimum (first min), then a local maximum, then a local minimum (second min)? No, the end is down, so after the second local maximum, it goes down, so there's no second local minimum. Wait, no, the graph's shape: for a degree 4 polynomial with leading coefficient negative, the graph will have a "W" shape (two minima and one maximum) or "M" shape (two maxima and one minimum)? Wait, no, leading coefficient negative: the graph opens downward (like a parabola but degree 4). So it would have a "W" shape? No, a degree 4 polynomial with negative leading coefficient: as x→±∞, f(x)→-∞, so the graph is like a "valley" with two peaks? No, wait, no: for even degree, negative leading coefficient: the graph is symmetric in end behavior (both down), and the number of minima and maxima: if it's a quartic (degree 4) with negative leading coefficient, it can have 3 turning points: 2 minima and 1 maximum, or 2 maxima and 1 minimum. Wait, the graph in the problem: let's look at the given graph. The graph has a peak (local max), then a valley (local min), then a peak (local max), then down. Wait, that's 2 maxima and 1 minimum. So relative minima: 1.

So putting it together:

Degree: even (4, since 3 turning points, n - 1 = 3 ⇒ n = 4), leading coefficient: negative (since ends down), different real zeros: 2 (crosses x-axis twice), relative minima: 1.

Wait, but maybe the degree is 4 (even), leading coefficient negative, 2 different real zeros, 1 relative minimum.

Step2: Confirm each part

- Degree: even (since both ends down), and number of turning points is 3, so degree 4 (n - 1 = 3 ⇒ n = 4).
- Leading coefficient: negative (ends down, even degree).
- Different real zeros: the graph crosses the x-axis at two distinct points (left and right), so 2.
- Relative minima: the graph has one local minimum (the middle valley).

Answer:

The degree of \( f(x) \) is \(\boldsymbol{4}\) (even) and the leading coefficient is \(\boldsymbol{\text{negative}}\). There are \(\boldsymbol{2}\) different real zeros and \(\boldsymbol{1}\) relative minimums.