Question

Roller Coaster Crew
Ray and Kelsey have summer internships at an engineering firm. As part of their internship, they get to assist in the planning of a brand new roller coaster. For this assignment, you help Ray and Kelsey as they tackle the math behind some simple curves in the coaster's track.

Part A

The first part of Ray and Kelsey's roller coaster is a curved pattern that can be represented by a polynomial function.

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.

Answer 1

Answer:

Step-by-step explanation:

In general terms, an nth degree polynomial has n zeros, for example, a first-degree polynomial (a line) has 1 zero. Then a third-degree polynomial function  has 3 zeros, as Kelsey said. But Ray said the function has 4 intercepts, which are the 3 zeros (interceptions with the x-axis) and the y-intercept (every polynomial has 1 y-intercept), so he is also correct.

Answer 2

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

• No, Ray and Kelsey cannot be correct together, as 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 zeros and y-intercept of a function?

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

Thus, if 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 same (not distinct) (although for not same cases too, we can count them different).

Thus, No, Ray and Kelsey cannot be correct together, as 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 behaviors, we have to analyse output of polynomial as x becomes arbitrarily large or small. Taking limits, we get:

$lim_{x\rightarrow \infty}(g(x) = (x + 2)(x - 1)(x - 2)) \implies g(x) \rightarrow \infty\\\\lim_{x\rightarrow -\infty}(g(x) = (x + 2)(x - 1)(x - 2)) \implies g(x) \rightarrow -\infty$

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

Their graphs are attached below.

Thus, the solutions to the sub parts are:

• No, Ray and Kelsey cannot be correct together, as 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 here:

https://brainly.com/question/19508384