Question

homework assignment 3.3: power functions and polyno
score: 9/10 answered: 9/10
question 10
what is the least possible degree of the polynomial graphed above?
question help: video written example

Explanation:

Step1: Analyze the end - behavior of the polynomial graph

The end - behavior of a polynomial is determined by the leading term \(a_nx^n\), where \(n\) is the degree of the polynomial and \(a_n\) is the leading coefficient. For the given graph, as \(x\to+\infty\), \(y\to-\infty\) and as \(x\to-\infty\), \(y\to-\infty\). This means that the leading coefficient \(a_n\) is negative and the degree \(n\) of the polynomial is even (because for even - degree polynomials \(y = a_nx^n\), when \(a_n<0\), the graph goes down on both ends).

Step2: Analyze the number of turning points

The number of turning points of a polynomial function \(y = f(x)\) is at most \(n - 1\), where \(n\) is the degree of the polynomial. The given graph has a single turning point (a maximum point). If the number of turning points is \(T\), then \(T\leq n - 1\). Here, \(T = 1\), so \(n-1\geq1\), which implies \(n\geq2\). But we also know from the end - behavior that \(n\) is even. The smallest even integer greater than or equal to 2 is 2? Wait, no, wait. Wait, let's re - examine the graph. Wait, the graph crosses the \(x\) - axis at two points? Wait, no, looking at the graph, the graph touches or crosses the \(x\) - axis? Wait, the graph intersects the \(x\) - axis at two distinct points? Wait, no, the graph has a vertex (a maximum) and it crosses the \(x\) - axis at two points. Wait, but for a quadratic function (degree 2), the graph is a parabola, which has one turning point and two \(x\) - intercepts (if the discriminant is positive). But wait, the end - behavior: for a quadratic function \(y=ax^2 + bx + c\), if \(a<0\), as \(x\to\pm\infty\), \(y\to-\infty\), which matches the given graph. But wait, wait, the graph in the picture: let's count the number of \(x\) - intercepts. Wait, the graph crosses the \(x\) - axis at two points? Wait, no, maybe I made a mistake. Wait, the graph of a quadratic function (degree 2) has one turning point. But wait, the given graph: let's check the end - behavior again. Wait, as \(x\) approaches positive infinity, \(y\) approaches negative infinity, and as \(x\) approaches negative infinity, \(y\) approaches negative infinity. So the degree is even. The number of turning points: the graph has 1 turning point. The formula for the maximum number of turning points is \(n - 1\). So if \(n-1\geq1\), \(n\geq2\). But also, the graph intersects the \(x\) - axis at two points. For a quadratic function \(y = ax^2+bx + c\), the number of real roots is given by the discriminant \(\Delta=b^2 - 4ac\). If \(\Delta>0\), there are two distinct real roots. But wait, is the graph a quadratic? Wait, no, wait a second. Wait, the graph: let's look at the shape. Wait, a quadratic function (degree 2) has a parabola shape. But the given graph: wait, maybe I miscounted the turning points. Wait, no, the graph has one turning point (a maximum). But wait, the degree: let's think again. Wait, the end - behavior: both ends go down, so even degree. The number of turning points: 1. So \(n-1\geq1\) implies \(n\geq2\). But wait, is there a mistake here? Wait, no, the smallest even degree with at least 1 turning point is 2? Wait, no, wait, a quadratic function (degree 2) has exactly 1 turning point (when \(n = 2\), \(n-1=1\)). So the least possible degree is 2? Wait, no, wait, maybe I made a mistake. Wait, no, let's check the graph again. Wait, the graph: when \(x = 0\), \(y = 1\) (the y - intercept). The graph has a maximum at \(x = 0\) (since it's symmetric about the y - axis) and crosses the \(x\) - axis at two points. So this is a parabola, which is a quadratic function (degree 2). But wait, wait, the problem says "least possible degree". Wait, but maybe I am wrong. Wait, no, the end - behavior: for degree 2, even, leading coefficient negative. Turning points: 1, which is \(n - 1=1\) (since \(n = 2\)). So the least possible degree is 2? Wait, no, wait, no, wait a minute. Wait, the graph: let's see, the number of \(x\) - intercepts. If a polynomial has \(k\) distinct real roots, its degree \(n\geq k\). Here, the graph crosses the \(x\) - axis at two points, so \(n\geq2\). And since the end - behavior is even - degree and the number of turning points is 1 (\(n-1 = 1\) implies \(n = 2\)), the least possible degree is 2? Wait, no, wait, I think I made a mistake. Wait, no, the graph: wait, the shape. Wait, a quadratic function is a parabola, which has one turning point. The given graph has one turning point. So the degree should be 2? But wait, no, wait, the problem is about the least possible degree. Wait, maybe the graph is not a quadratic. Wait, no, let's recall the properties. The key points are:

1. End - behavior: both ends down, so even degree.
2. Number of turning points: 1, so \(n-1\geq1\) (so \(n\geq2\)).
3. Number of \(x\) - intercepts: 2, so \(n\geq2\).

The smallest even integer that satisfies \(n\geq2\) and \(n\) is even is \(n = 2\)? Wait, no, wait, no. Wait, a quadratic function (degree 2) has exactly one turning point and can have two \(x\) - intercepts (when discriminant is positive). And the end - behavior for \(a<0\) is down on both ends. So the least possible degree is 2? Wait, but I think I made a mistake. Wait, no, wait, the graph: let's look at the graph again. Wait, the user's graph: the parabola - like shape, with a maximum, crossing the \(x\) - axis at two points, symmetric about the y - axis. So it's a quadratic function, degree 2. But wait, no, wait, maybe the graph has a multiplicity of roots? Wait, no, the graph crosses the \(x\) - axis, so the roots are real and distinct. So the degree should be at least 2. And since it's a parabola (quadratic), the degree is 2. Wait, but I think I was wrong earlier. Wait, the correct answer is 2? Wait, no, wait, no. Wait, the end - behavior: for a quadratic function, when \(a<0\), as \(x\to\pm\infty\), \(y\to-\infty\), which matches. The number of turning points: 1, which is \(n - 1=1\) (so \(n = 2\)). The number of \(x\) - intercepts: 2, so \(n\geq2\). So the least possible degree is 2. Wait, but I think I made a mistake. Wait, no, let's check with an example. Let \(y=-x^2 + 1\). This is a quadratic function (degree 2), leading coefficient - 1 (negative), end - behavior: as \(x\to\pm\infty\), \(y\to-\infty\). It has a maximum at \(x = 0\) (one turning point), and \(x\) - intercepts at \(x=\pm1\). Which matches the given graph's characteristics (symmetric about y - axis, maximum at y - axis, two \(x\) - intercepts, end - behavior down on both ends). So the least possible degree is 2. Wait, but I think I was confused earlier. So the least possible degree is 2.

Answer:

2