Question

1. Find the serves of the
4002 -
quadudic polynomial​

Answer 1

Answer: F(x) = 3x^2 +2x + 1

Step-by-step explanation:

There are several ways of doing this. I think the easiest is by setting up a system of equations. The general quadratic is F(x) = ax^2 +bx + c. If a point is on a graph, it means it satisfies the equation. So,

F(-2) = 9    or a(-2)^2 + b(-2) + c = 9       or 4a - 2b + c = 9

F(0) = 1                                                                           c = 1

F(2) = 17                                                         4a + 2b +c = 17

Replacing c with 1 in the first and third equations we get,

4a - 2b = 8

4a +2b = 16

Already set up for elimination, so add to get

8a = 24     or a=3

Substitute back

4a + 2b + c = 17

4(3) + 2b +1 = 17

2b +13 =17     so b= 2

So F(x) = 3x^2 +2x + 1 Always check. See if your 3 points satisfy this. I did and it works.