Applying Polynomial Functions to Real-World Situations

To order the given functions according to their degrees, we first need to determine their degrees. Let's consider each function separately.

Function f(x)

From the given graph, we can see that the function crosses the x-axis four times. In other words, the graph has four x-intercepts.

This suggests that the degree of the polynomial function f(x) is at least 4. Since Ignacio's teacher said that f(x) has the least possible degree that it can have, we can conclude that the degree of f(x) is indeed 4.

Function g(x)

Given a function rule, the degree of the polynomial is equal to the greatest exponent. Let's consider the given function rule again. g(x)=x^3-2x^2+3x-8 The greatest exponent in the function is 3, so the degree of g(x) is 3.

Function h(x)

The function h(x) is given as a table of values. Notice that the x-values are equally spaced, meaning that we can use the properties of finite differences to determine the degree of the polynomial function that best fits the data set.

Let's find the first differences by subtracting each pair of consecutive function values.

Since the first differences are not constant, let's proceed to finding the second differences.

The second differences are constant and different from zero. Therefore, the degree of the polynomial h(x) is 2.

Function k(x)

The function k(x) is also given as a table of values, but notice that the x-values are not equally spaced. This means that we cannot proceed as we did with h(x). However, no pair of points lie on a vertical line — all the x-values are different. Therefore, we can use The (n+1) Point Principle.

The (n+1) Point Principle |- Given a set of n+1 points in the coordinate plane, where no pair of points lies on a vertical line, there is a unique polynomial of degree at most n that fits the points perfectly.

In our case, we were given 5+1=6 different points. Therefore, there is a unique polynomial of degree at most 5 that fits the points perfectly. Since the teacher said that k(x) has the greatest possible degree that it can have, we can conclude that the degree of k(x) is 5.

Conclusion

Now that we know the degree of each polynomial function, we can order them according to their degrees, from least to greatest.

The first thing we notice from the given information and graph is that it has three x-intercepts.

Each x-intercept represents a zero of the function. By the Factor Theorem, each zero gives us a factor the polynomial function.

Since the function is cubic, it can be written in factored form as a constant a multiplied by the three factors we identified. f(x) = a(x+4)(x-1)(x-5) Let's use the y-intercept of the graph to determine the value of a. The y-intercept is (0,40),so we have that f(0)=40. If we evaluate f(x) at x=0, we can solve the resulting equation for a.

Now that we know the value of a, we can substitute it into the factored form. We can then perform the required multiplications to write the function rule in standard form.

Since we were given only 4 data values, we know that by the (n+1) Point Principle, a polynomial of degree at most 3 exists that fits the points perfectly. In other words, we are looking for a cubic polynomial. P(x) = ax^3 + bx^2 + cx + d To find the values of the coefficients, we can substitute the study times for x and the test scores for f(x). This will generate a system of four equations. 92 = a( 12)^3 + b( 12)^2 + c( 12) + d 32 = a( 16)^3 + b( 16)^2 + c( 16) + d 26 = a( 18)^3 + b( 18)^2 + c( 18) + d 50 = a( 19)^3 + b( 19)^2 + c( 19) + d Let's perform the required operations and rearrange the system so that the results are on the right-hand side. 1728a + 144b + 12c + d = 92 4096a + 256b + 16c + d = 32 5832a + 324b + 18c + d = 26 6859a + 361b + 19c + d = 50 Since the system is big, solving it by elimination or substitution can be messy. Instead, let's use a graphing calculator. First, we input the system of equations as a matrix. To do so, we press 2ND, then x^(-1), and scroll to the right to reach the EDIT menu.

In that menu, we can select any of the matrices. After that, the calculator shows a window where we can enter the elements of our system. To input a system of four equations with four variables, a 4* 5 matrix will be needed. The last column stores the results on the right-hand side of each equation.

To solve the system of equations, we press 2ND and x^(-1) again. However, this time, we will select the MATH menu and scroll down until we select the rref( option.

Finally, we input the matrix [A] into the rref( function. To do so, push 2ND and x^(-1) to bring up all available matrices. In the NAMES menu, choose matrix [A].

The solutions to the system of equation are the numbers in the last column. For example, the value of a is the first number, the value of b is the second one, and so on. a = 1 b=-44 c=625 d=-2800 Finally, let's substitute these values into the cubic polynomial. P(x) = x^3 - 44x^2 + 625x - 2800 The graph of this function looks as follows.