The following set of points belong to a specific function:

{(-3,0)(-2,4), (-1,0), (0,-6),(1,-8), (2,0),(3,24)}
Based on the set of points answer the following questions:
a) What type of function does it produce? Justify your answer.

b) Write an equation

Through n +1 points passing polynomial of degree n.

In this case you have 7 points.

Polynomial will be degree 6.

Write the polynomial in this form :

a * x ^ 6 + b * x ^ 5 + c * x ^ 4 + d * x ^ 3 + e * x ^ 2 + f *x + g = y

Now put values of x an y in this equation.

a * ( - 3 ) ^ 6 + b * ( - 3 ) ^ 5 + c * ( - 3 ) ^ 4 + d * ( - 3 ) ^ 3 + e * ( - 3 ) ^ 2 + f * ( - 3 ) + g = 0

a * ( - 2 ) ^ 6 + b * ( - 2 ) ^ 5 + c * ( - 2 ) ^ 4 + d * ( - 2 ) ^ 3 + e * ( - 2 ) ^ 2 + f * ( - 2 ) + g = 4

a * ( - 1 ) ^ 6 + b * ( - 1 ) ^ 5 + c * ( - 1 ) ^ 4 + d * ( - 1 ) ^ 3 + e * ( - 1 ) ^ 2 + f * ( - 1 ) + g = 0

a * 0 ^ 6 + b * 0 ^ 5 + c * 0 ^ 4 + d * 0 ^ 3 + e * 0 ^ 2 + f * 0 + g = - 6

a * 1 ^ 6 + b * 1 ^ 5 + c * 1 ^ 4 + d * 1 ^ 3 + e * 1 ^ 2 + f * 1 + g = - 8

a * 1 ^ 6 + b * 1 ^ 5 + c * 1 ^ 4 + d * 1 ^ 3 + e * 1 ^ 2 + f * 1 + g = - 8

a * 2 ^ 6 + b * 2 ^ 5 + c * 2 ^ 4 + d * 2 ^ 3 + e * 2 ^ 2 + f * 2 + g = 0

a * 3 ^ 6 + b * 3 ^ 5 + c * 3 ^ 4 + d * 3 ^ 3 + e * 3 ^ 2 + f * 3 + g = 24

OR

729 a - 243 b + 81 c - 27 d + 9 e - 3 f + g = 0

64 a - 32 b + 16 c - 8 d + 4 e - 2 f + g = 4

a - b + c - d + e - f + g = 0

0 + 0 + 0 + 0 + 0 + 0 + g = - 6

a + b + c + d + e + f + g = - 8

64 a + 32 b + 16 c + 8 d + 4 e + 2 f + g = 0

729 a + 243 b + 81 c + 27 d + 9 e + 3 f + g = 24

Solutions of this system of equations are :

a = 0

b = 0

c = 0

d = 1

e = 2

f = - 5

g = - 6

Your Polynomial :

y = 0 * x ^ 6 + 0 * x ^ 5 + 0 * x ^ 4 + 1 * x ^ 3 + 2 * x ^ 2 + ( - 5 ) * x + ( - 6 )

y = x ^ 3 + 2 x ^ 2 - 5 x - 6

That is cubic parabola.

If you want to see graph go on :

wolframalpha dot com

When page be open in rectangle type :

intepolating polynomial {(-3,0),(-2,4), (-1,0), (0,-6),(1,-8), (2,0),(3,24)}

and click option =

You will see graph

a) The given set of points does not represent a linear function because there is no constant rate of change between the x-values and the corresponding y-values. In a linear function, the y-values change by a fixed amount for each unit change in x. However, in this case, the values of y are changing differently for different values of x. Therefore, the given set of points does not produce a linear function.

b) To find the equation for the function, we can try to identify any patterns or relationships in the data. Looking at the x-values and their corresponding y-values, it appears that the y-values are increasing as the absolute value of x increases.

One possible equation that can represent this relationship is a quadratic equation of the form y = ax^2 + bx + c. Let's substitute the given points into this equation to determine the values of a, b, and c.

For (-3, 0): 0 = a(-3)^2 + b(-3) + c
0 = 9a - 3b + c --------- (1)

For (-2, 4): 4 = a(-2)^2 + b(-2) + c
4 = 4a - 2b + c --------- (2)

For (-1, 0): 0 = a(-1)^2 + b(-1) + c
0 = a - b + c --------- (3)

For (0, -6): -6 = a(0)^2 + b(0) + c
-6 = c --------- (4)

For (1, -8): -8 = a(1)^2 + b(1) + c
-8 = a + b + c --------- (5)

For (2, 0): 0 = a(2)^2 + b(2) + c
0 = 4a + 2b + c --------- (6)

For (3, 24): 24 = a(3)^2 + b(3) + c
24 = 9a + 3b + c --------- (7)

We now have a system of equations (1)-(7) that we can solve to find the values of a, b, and c.

By using equations (4) and (6), we find that c = -6 and 4a + 2b - 6 = 0, which simplifies to 2a + b - 3 = 0.

We can also solve equations (3) and (5) to obtain a - b + c = 0 and a + b + c = -8. Subtracting these two equations, we get 2b = -8, which leads to b = -4. Substituting this value of b into equation 3, we find that a - (-4) - 6 = 0, which simplifies to a + 4 = 6. Therefore, a = 2.

By substituting the values of a, b, and c into the equation y = ax^2 + bx + c, we obtain the equation y = 2x^2 - 4x - 6 as an equation that represents the given set of points.

a) To determine the type of function, we need to analyze the pattern or relationship between the given set of points. One way to do this is to create a table with the x-values and corresponding y-values:

x | y
--------------
-3 | 0
-2 | 4
-1 | 0
0 | -6
1 | -8
2 | 0
3 | 24

By observing the y-values, we can see that they increase, then decrease, and then increase again. This indicates that the function is not strictly increasing or strictly decreasing. Additionally, the y-values go from positive to negative and back to positive, suggesting that the function may be symmetric.

Given these observations, the set of points appears to produce a function that is a quadratic function. Quadratic functions are known for their u-shape and can be symmetric about the axis of symmetry.

b) Now let's find the equation for the quadratic function. We can use the general form of a quadratic function:

y = ax^2 + bx + c

To determine the values of a, b, and c, we can substitute the x and y values from one of the given points into the equation. Let's use the point (-2, 4), for example.

4 = a(-2)^2 + b(-2) + c

Simplifying this equation:

4 = 4a - 2b + c ... (Equation 1)

Next, let's use another point to create another equation. Using the point (1, -8):

-8 = a(1)^2 + b(1) + c

Simplifying:

-8 = a + b + c ... (Equation 2)

Now we have a system of two equations (Equation 1 and Equation 2) with three unknowns (a, b, and c). To solve this system, we can substitute Equation 2 into Equation 1:

4 = 4a - 2b + (-8 - a - b)

Simplifying:

4 = 3a - 3b - 8

Rearranging:

3a - 3b = 12

Dividing both sides by 3:

a - b = 4 ... (Equation 3)

Now we can solve Equation 3 simultaneously with Equation 2 to find the values of a, b, and c.

Using Equation 3 and Equation 2:

(a - b) + b + c = 4 + (-8)

Simplifying:

a + c = -4 ... (Equation 4)

Now we have two equations (Equation 2 and Equation 4) with two unknowns (a and c). To eliminate c, subtract Equation 4 from Equation 2:

(a + b + c) - (a + c) = -8 - (-4)

Simplifying:

b = -4

Now we can substitute the found values of b into Equation 2 to solve for a:

-8 = a + (-4) + c

Simplifying:

-8 = a - 4 + c

Rearranging:

a + c = -4 ... (Equation 5)

We can observe that Equation 4 and Equation 5 are identical. Therefore, there are infinitely many solutions for a and c.

In conclusion, the equation for the quadratic function cannot be determined solely based on the given set of points. Further information or additional points are required to find a unique equation.