6.9 Modeling with Polynomial Functions

Ex: Write the cubic function whose graph goes through the points (-2,0), (0,2), (1,0), and (3,0). The 3 x-intercepts (-2,0), (1,0), and (3,0) will give you the 3 zeros of the cubic. They will also tell you 3 factors to use f(x)=a(x+2)(x-1)(x-3). Then use the 4th point as x & f(x) values. 2=a(0+2)(0-1)(0-3) Now solve for a! 2=6a so, a=1/3 Answer: f(x)=1/3(x+2)(x-1)(x-3)

Ex: An eqn. for a polynomial function is f(n)=2n3+n2+2n+1 Ex: An eqn. for a polynomial function is f(n)=2n3+n2+2n+1. Show that this function has constant 3rd order differences. (you check the 3rd order diffs. because it’s a degree 3 polynomial) First, write out the first several values; or find f(1), f(2), f(3),…, f(6). f(1)=6 f(2)=25 f(3)=70 f(4)=153 f(5)=286 f(6)=481 Now subtract #s! (left from right) 1st diffs. 19 45 83 133 195 2nd diffs 26 38 50 62 3rd diffs. 12 12 12 ** This is called using finite differences.

Ex:The values of a polynomial function for six consecutive whole numbers are given below. Write a polynomial function for f(n). f(1)= -2, f(2)=2, f(3)=12, f(4)=28, f(5)=50, andf(6)=78 First, find finite differences. (Stop when the same number repeats all the way across!) 4 10 16 22 28 6 6 6 6 The 2nd differences are now a constant # across. This means the function will be a quadratic. (degree 2) So, use f(n)=an2+bn+c. Since you must find a, b, & c, you will need to make 3 eqns. with these 3 variables using the first 3 known values of the function.

Use an2+bn+c=f(n) & f(1)= -2, f(2)=2, f(3)=12 to write 3 equations. a(1)2+b(1)+c= -2 a+b+c= -2 a(2)2+b(2)+c=2 4a+2b+c=2 a(3)2+b(3)+c=12 9a+3b+c=12 ** Look familiar? It should! ** * Use inverse matrices to solve for a, b, &c! * This means the quadratic is f(n)=3n2-5n+0 or f(n)=3n2-5n

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