Answer:
(a)Degree 3
(b)[tex]f(x)=x^3+2x^2-7x+3 .[/tex]
Step-by-step explanation:
The function represented by the set of points: {(-3,15)(-2,17), (-1,11), (0,3),(1,-1), (2,5),(3,27)} has 2 turning points when plotted on a graph.
(a)Now, we know that 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 [tex]f(x)=ax^3+bx^2+cx+d .[/tex] where d is the y-intercept.
From the set of values, the y-intercept, d=3
Therefore, our polynomial is of the form:
[tex]f(x)=ax^3+bx^2+cx+3 .[/tex]
[tex]\text{At } (-3,15), 15=a(-3)^3+b(-3)^2+c(-3)+3 \implies -27a+9b-3c=12\\\text{At } (-2,17), 17=a(-2)^3+b(-2)^2+c(-2)+3 \implies -8a+4b-2c=14\\\text{At } (-1,11), 11=a(-1)^3+b(-1)^2+c(-1)+3 \implies -a+b-c=8[/tex]
Solving the three resulting equations simultaneously use a calculator), we obtain:
[tex]a=1, b=2, c=-7[/tex]
Therefore, an equation of this function is:
[tex]f(x)=x^3+2x^2-7x+3 .[/tex]
