Question

Find a polynomial of the form

f(x) = ax3 + bx2 + cx + d

such that f(0) = −3, f(1) = 2, f(3) = 5, and f(4) = 0. (A graphing calculator is recommended.)

Answer 1

Answer:

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

Step-by-step explanation:

We are given that

$f(x)=ax^3+bx^2+cx+d$

f(0)=-3,f(1)=2,f(3)=5 and f(4)=0

We have to find the polynomial

Substitute the value x=0 then ,we get

f(0)=d=-3

Substitute x=1 then we get

$a+b+c-3=2$

$a+b+c=2+3=5$

$a+b+c=5$ (equation I)

Substitute x=3 then we get

$a(3)^3+b(3)^2+c(3)-3=5$

$27a+9b+3c=5+3=8$

$27a+9b+3c=8$  (Equation II)

Substitute x=4 then we get

$a(4)^3+b(4)^2+c(4)-3=0$

$64a+16b+4c=3$ (Equation III)

Equation I multiply by 3 then subtract from equation II

$24a+6b=-7$ (Equation IV)

Equation II multiply by 4 and equation III multiply by 3 and subtract equation II from III

$84a+12b=-23$ (Equation V)

Equation IV multiply by 2 and then subtract from equation V

$36a=-9$

$a=-\frac{9}{36}$

$a=-\frac{1}{4}$

Substitute the value of a in equation IV then we get

$24(-\frac{1}{4})+6b=-7$

$-6+6b=-7$

$6b=-7+6=-1$

$b=-\frac{1}{6}$

Substitute the value of b in equation I then we get

$-\frac{1}{4}-\frac{1}{6}+ c=5$

$-\frac{5}{12}+c=5$

$c=5+\frac{5}{12}=\frac{65}{12}$

Substitute the values then we get

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$