Answer:
(a)Degree 3
(b)
![f(x)=x^{3}+2 x^{2}-7 x+3](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E%7B3%7D%2B2%20x%5E%7B2%7D-7%20x%2B3)
Step-by-step explanation:
(a)Given the function represented by the set of points: {(-3,15)(-2,17), (-1,11), (0,3),(1,-1), (2,5),(3,27)}.
To determine the degree of the polynomial of the function, we plot the function on a graph.
From the graph, the function has 2 turning points.
The maximum number of turning points of a polynomial function is always one less than the degree of the function.
Therefore, the polynomial has a degree of 3
.
(b)A cubic function is one in the form where d is the y-intercept.
A cubic function is one in the form
where d is the y-intercept.
- From the point (0,3) the y-intercept, d=3
Therefore, our polynomial is of the form:
![\begin{array}{l}f(x)=a x^{3}+b x^{2}+c x+3 \\\text {At }(-3,15), 15=a(-3)^{3}+b(-3)^{2}+c(-3)+3 \implies -27 a+9 b-3 c=12 \\\text {At }(-2,17), 17=a(-2)^{3}+b(-2)^{2}+c(-2)+3 \Longrightarrow -8 a+4 b-2 c=14 \\\text {At }(-1,11), 11=a(-1)^{3}+b(-1)^{2}+c(-1)+3 \Longrightarrow -a+b-c=8\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7Df%28x%29%3Da%20x%5E%7B3%7D%2Bb%20x%5E%7B2%7D%2Bc%20x%2B3%20%5C%5C%5Ctext%20%7BAt%20%7D%28-3%2C15%29%2C%2015%3Da%28-3%29%5E%7B3%7D%2Bb%28-3%29%5E%7B2%7D%2Bc%28-3%29%2B3%20%5Cimplies%20-27%20a%2B9%20b-3%20c%3D12%20%5C%5C%5Ctext%20%7BAt%20%7D%28-2%2C17%29%2C%2017%3Da%28-2%29%5E%7B3%7D%2Bb%28-2%29%5E%7B2%7D%2Bc%28-2%29%2B3%20%5CLongrightarrow%20-8%20a%2B4%20b-2%20c%3D14%20%5C%5C%5Ctext%20%7BAt%20%7D%28-1%2C11%29%2C%2011%3Da%28-1%29%5E%7B3%7D%2Bb%28-1%29%5E%7B2%7D%2Bc%28-1%29%2B3%20%5CLongrightarrow%20-a%2Bb-c%3D8%5Cend%7Barray%7D)
Solving the three resulting equations simultaneously (using a calculator), we obtain:
a=1, b=2, c=-7
Therefore, the equation for this function is:
![f(x)=x^{3}+2 x^{2}-7 x+3](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E%7B3%7D%2B2%20x%5E%7B2%7D-7%20x%2B3)