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 $\square$ different real zeros and $\square$ relative minimums.

Explanation:

Step1: Determine the degree of the polynomial

The end - behavior of a polynomial function is determined by the leading term \(a_nx^n\), where \(n\) is the degree and \(a_n\) is the leading coefficient. For a polynomial, if the ends of the graph both go up (as \(x ightarrow+\infty\) and \(x ightarrow - \infty\)), the degree \(n\) is even. Also, the number of turning points (local maxima and minima) of a polynomial function \(y = f(x)\) is at most \(n - 1\). Looking at the graph, we can see that there are 3 turning points (2 minima and 1 maximum). Since the number of turning points \(T\leq n - 1\), and \(T = 3\), then \(n-1\geq3\), so \(n\geq4\). Also, the end - behavior (both ends up) implies that the degree is even. Let's check the number of real zeros (x - intercepts). The graph crosses the x - axis at 2 points? Wait, no, looking at the graph, it crosses the x - axis at 2 points? Wait, no, the graph as shown: let's count the x - intercepts. Wait, the graph crosses the x - axis at two points? Wait, no, actually, when we look at the graph, the polynomial has a degree of 4? Wait, no, the number of turning points: for a degree \(n\) polynomial, the maximum number of turning points is \(n - 1\). Here, we have 3 turning points (2 minima and 1 maximum), so \(n-1 = 3\) implies \(n = 4\)? Wait, no, 3 turning points mean \(n\geq4\). But the end - behavior: both ends go up, so the leading coefficient is positive and the degree is even. Let's confirm:

Degree: The graph has 3 turning points (local max and mins). The formula for the maximum number of turning points is \(n - 1\), so if there are 3 turning points, \(n-1\geq3\), so \(n\geq4\). Also, the end - behavior (as \(x ightarrow\pm\infty\), \(f(x) ightarrow+\infty\)) implies that the leading coefficient is positive and the degree is even. Let's assume the degree is 4 (a quartic polynomial) because with 3 turning points, \(n = 4\) (since \(n-1 = 3\) when \(n = 4\)).

Step2: Determine the leading coefficient

Since as \(x ightarrow+\infty\) and \(x ightarrow-\infty\), the graph of the polynomial goes to \(+\infty\), for a polynomial \(f(x)=a_nx^n+\cdots+a_0\), when \(n\) is even, if \(a_n>0\), then \(f(x) ightarrow+\infty\) as \(x ightarrow\pm\infty\). So the leading coefficient is positive.

Step3: Determine the number of real zeros

The real zeros of a polynomial are the x - intercepts (where the graph crosses or touches the x - axis). Looking at the graph, we can see that the graph crosses the x - axis at 2 points? Wait, no, wait the graph: let's look again. The graph crosses the x - axis at two distinct points? Wait, no, actually, the graph as drawn: it crosses the x - axis at two points? Wait, no, maybe I made a mistake. Wait, the graph: let's see, the left side crosses the x - axis, then comes down, has a local max, then a local min, then crosses the x - axis again. Wait, so that's 2 real zeros? Wait, no, wait the graph: maybe it's two x - intercepts? Wait, no, the graph shown: let's count the number of times it intersects the x - axis. The graph crosses the x - axis at 2 points? Wait, no, wait the graph: when we look at the graph, it crosses the x - axis at two points? Wait, no, maybe I misread. Wait, the graph: starts from the top left (up), crosses the x - axis, goes down, has a local minimum, then a local maximum, then goes down again, has a local minimum, then goes up and crosses the x - axis again. So that's 2 x - intercepts? Wait, no, that's two real zeros? Wait, no, wait the number of x - intercepts: the graph crosses the x - axis at two points, so two real zeros? Wait, no, wait the graph: let's see, the left crossing, then the right crossing. So two real zeros? Wait, no, maybe I made a mistake. Wait, the degree is 4, so the number of real zeros can be 2 (with multiplicity, but the question says "different real zeros"). So different real zeros: 2? Wait, no, wait the graph: maybe it's two x - intercepts, so two different real zeros.

Step4: Determine the number of relative minimums

Relative minimums are the "valleys" in the graph (local minima). Looking at the graph, we can see that there are 2 relative minimums (the two lowest points in the graph).

Answer:

The degree of \(f(x)\) is \(\boldsymbol{4}\) (even, since both ends go up) and the leading coefficient is \(\boldsymbol{\text{positive}}\). There are \(\boldsymbol{2}\) different real zeros and \(\boldsymbol{2}\) relative minimums.

Wait, let's re - check the degree. The number of turning points: for a degree \(n\) polynomial, the number of turning points \(T\) satisfies \(T\leq n - 1\). Here, we have 3 turning points (2 minima and 1 maximum), so \(n-1\geq3\), so \(n\geq4\). The end - behavior (both ends up) implies \(n\) is even. So the smallest even \(n\) with \(n - 1\geq3\) is \(n = 4\) (since \(4-1 = 3\)). So degree is 4. Leading coefficient: positive (because both ends go up and degree is even). Number of real zeros: the graph crosses the x - axis at 2 points, so 2 different real zeros. Number of relative minimums: looking at the graph, there are 2 relative minimums (the two low points).