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Ray and Kelsey are working to graph a third-degree polynomial function that represents the first pattern in the coaster plan. Ray says the third-degree polynomial has four intercepts. Kelsey argues the function can have as many as three zeros only. Is there a way for the both of them to be correct? Explain your answer.

Kelsey has a list of possible functions. Pick one of the g(x) functions below and then describe to Kelsey the key features of g(x), including the end behavior, y-intercept, and zeros.
g(x) = (x + 2)(x − 1)(x − 2)
g(x) = (x + 3)(x + 2)(x − 3)
g(x) = (x + 2)(x − 2)(x − 3)
g(x) = (x + 5)(x + 2)(x − 5)
g(x) = (x + 7)(x + 1)(x − 1)

Create a graph of the polynomial function you selected from Question 2.

Part B

The second part of the new coaster is a parabola.

Ray needs help creating the second part of the coaster. Create a unique parabola in the pattern f(x) = (x − a)(x − b). Describe the direction of the parabola and determine the y-intercept and zeros.

Create a graph of the polynomial function you created in Question 4.

Part C

Now that the curve pieces are determined, use those pieces as sections of a complete coaster. By hand or by using a drawing program, sketch a design of Ray and Kelsey's coaster that includes the shape of the g(x) and f(x) functions that you chose in the Parts A and B. You do not have to include the coordinate plane. You may arrange the functions in any order you choose, but label each section of the graph with the corresponding function for your instructor to view.

Respuesta :

Answer:

Part A

Please find attached the graph of the function g(x) = (x + 2)·(x - 1)·(x - 2) created with  MS Excel

Part B

Please find attached the parabola created with MS Excel

The shape of the parabola = cup shape

y-intercept is y = 4

The zeros, is x = 2

Part C

Please find attached the graph of the combined coaster created with MS Excel

Step-by-step explanation:

Part A

The maximum number of x-intercepts of a third degree polynomial is between 1 and 3, however, the third degree polynomial can have a y-intercept to make the number of intercepts = 3 + 1 = 4 intercept

a) For the function, g(x) = (x + 2)·(x - 1)·(x - 2), we have;

b) The x-intercepts which are the zeros of the function are;

x = -2, x = 1, and x = 2

The end behavior are;

When x → -∞, y → -∞, x → ∞, y → ∞

The y-intercept is y = (0 + 2)·(0 - 1)·(0 - 2) = 4

The y-intercept is y = 4

(ii) Please find attached the graph of the function created with MS Excel

Part B

The form of the parabola is f(x) = (x - a)·(x - b)

For a y-intercept of 4, we have;

f(x) = (x - 2)·(x - 2) = x² - 4·x + 4

Given that the coefficient of x² is negative, we have;

The parabola is opened up (cup shaped)

The y-intercept = 4

The zero is x = 2

Part C

Please find attached the combined coaster created with MS Excel

Ver imagen oeerivona
Ver imagen oeerivona
Ver imagen oeerivona

Zeros are the interception of the curve on the x-axis or input axis. The solutions to the subparts are:

  • No, Ray and Kelsey cannot be correct together, as a three-degree polynomial can have at max three intercepts.
  • Part A: The considered function is g(x) = (x + 2)(x − 1)(x − 2) . It  rises from bottom left to top right on its ends. Its zeros are x = -2, x = 1, and x = 2. It's y-intercept is at y = 4
  • Part B: The y-intercept of the parabola is on . Its zeros are x = 1, x = 2. It's y-intercept is at y = 2

How to find the zeros and y-intercept of a function?

  • Zeros or x-intercepts =Intersection points of the considered function's curve on the input axis(usually represented by values of x)
  • y-intercept = Intersection points  of the considered function's curve on the output axis
  • On the x-axis, the value of the y-coordinate is 0,
  • On the y-axis, the value of the x-coordinate is 0.

Thus, if the function is y=f(x) then putting y = 0 gives zeros of the function. And putting x = 0 gives the y-intercept of the function. (if any, for both cases).

How many zeros are possible for n order polynomial?

At max n distinct real zeros are possible for n order polynomial. It is "at max" since sometimes, some or all zeros may be imaginary, or sometimes some or all zeros may be the same (not distinct) (although for not the same cases too, we can count them different).

Thus, No, Ray and Kelsey cannot be correct together, as a three-degree polynomial can have at max three intercepts. (assuming intercept is denoting x-intercepts here)

Part A: Let we choose the function g(x) = (x + 2)(x − 1)(x − 2)

For end behaviours, we have to analyse the output of polynomial as x becomes arbitrarily large or small. Taking limits, we get:

[tex]lim_{x \rightarrow \infty}(g(x)=(x+2)(x-1)(x-2)) \implies g(x) \rightarrow \infty\\lim_{x \rightarrow -\infty}(g(x)=(x+2)(x-1)(x-2)) \implies g(x) \rightarrow -\infty[/tex]

Thus, the function's graph rises from bottom left to top right on its ends (which are not finite points but arbitrarily ahead or behind).

  • Finding zeros:

Putting y = g(x) = 0, we get:

g(x) = (x + 2)(x − 1)(x − 2) = 0

This output  must've become zero because of some or all factors evaluating:

(x+2) = 0, (x-1) = 0, (x-2) = 0

or

x = -2, x = 1, x = 2 (3 zeros).

  • Finding y-intercepts:

Putting x = 0, we get:

g(x) = (x + 2)(x − 1)(x − 2)

g(0) = 2(-1)(-2) = 4

Thus, the y-intercept of the considered function is at y = 4

Part B: Consider a = 1, b = 2 in  f(x) = (x − a)(x − b)

Then, we get: f(x) = (x-1)(x-2)

  • Finding zeros:

Putting y = f(x) = 0, we get:

f(x) = (x -1)(x-2) = 0

x = 1, x = 2 (3 zeros).

  • Finding y-intercepts:

Putting x = 0, we get:

f(x) = (x − 1)(x − 2)

f(0) = (-1)(-2) = 2

Thus, the y-intercept of the considered function is at y = 2

Their graphs are attached below.

Thus, the solutions to the subparts are:

  • No, Ray and Kelsey cannot be correct together, as a three-degree polynomial can have at max three intercepts.
  • Part A: The considered function is g(x) = (x + 2)(x − 1)(x − 2) . It  rises from bottom left to top right on its ends. Its zeros are x = -2, x = 1, and x = 2. It's y-intercept is at y = 4
  • Part B: The y-intercept of the parabola is on . Its zeros are x = 1, x = 2. It's y-intercept is at y = 2

Learn more about Zeros of a polynomial function:

https://brainly.com/question/19508384

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Ver imagen ap8997154

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