Question

Find the polynomial of the lowest possible degree which assumes the values 3, 12, 15, 21 when x has the values 3, 2, 1, 1, respectively. Hence find f(0).

Answer 1

Answer:

f(4)answer themself Hayek

Answer 2

The polynomial of the lowest possible degree is $f(x)=x^3-9x^2+17x+6$ and value of f(0) is 6 .

Step 1: Find the polynomial.

Given points - 3,12,15,-21

As we have 4 points and polynomial has a form y = ∑$^n _k_=_0$ $c_k$×$x^k$ , we know polynomial of third degree will satisfy this:

$a_3x^3+b_3x^2+c_3x+d_3=y$

This gives us 4 equations if we find values for x and y,

$27a_3+9b_3+3c_3+d_3=3$

$8a_3+4b_3+2c_3+d_3=12$

$a_3+b_3+c_3+d_3=15$

$-a_3+b_3-c_3+d_3=-21$

After solving this equation we get,

$a_3=1, b_3=-9, c_3=17, d_3=6$

so, $f(x)=x^3-9x^2+17x+6$

Step 2: Find value of f(0) .

$f(x)=x^3-9x^2+17x+6$

$f(0) = (0)^3 - 9(0)^2 +17(0) +6$

$f(0)= 6$