Answer:
(a)Degree 3
(b) [tex]f(x)=x^{3}+2 x^{2}-7 x+3[/tex]
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 [tex]f(x)=a x^{3}+b x^{2}+c x+d[/tex] where d is the y-intercept.
Therefore, our polynomial is of the form:
[tex]\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}[/tex]
Solving the three resulting equations simultaneously (using a calculator), we obtain:
a=1, b=2, c=-7
Therefore, the equation for this function is:
[tex]f(x)=x^{3}+2 x^{2}-7 x+3[/tex]
