f(x) = x² + bx + c
a = 1, b = ?, c = ?
a)
Z postaci iloczynowej:
f(x) = (x - x₁)(x - x₂)
f(x) = (x - 3)(x + 4) = x² + 4x - 3x - 12
f(x) = x² + x - 12
b = 1
c = -12
Lub
x = 3 i x = -6
f(3) = 0 i f(-6) = 0
3² + b·3 + c = 0
(-4)² + b·(-4) + c = 0
9 + 3b + c = 0
16 - 4b + c = 0 |·(-1)
9 + 3b + c = 0
-16 + 4b - c = 0
------------------------ +
-7 + 7b = 0
-7b = -7 /:(-7)
b = 1
c = -9 - 3b = -9 - 3·1
c = -12
Odp. b = 1, c = -12
[tex]b)\\\\W= (p, q)\\\\W = (2, -6) \ \ \rightarrow \ \ p = 2, \ q = -6\\\\f(x) = a(x-p)^{2}+a \ - \ postac \ kanoniczna\\\\a = 1\\\\f(x) =(x-2)^{2}-6 = x^{2}-4x+4 - 6\\\\f(x) = x^{2}-4x-2\\\\Odp. \ b = -4, \ c = -2[/tex]
