Znajdź wielomian f(x) drugiego stopnia taki że f(1)=-4, f(-1)=10, f(-2)=23 Według jakiego wzoru trzeba to zrobić?
f(x) = ax²+bx+c
f(1) = -4, f(-1) = 10, f(-2) = 23
-4 = a·1²+b·1+c
10 = a·(-1)²+b·(-1)+c
23 = a·(-2)²+b·(-2)+c
a+b+c = -4 => c = -4-a-b
a-b+c = 10
4a-2b+c = 23
a-b-4-a-b = 10
4a-2b-4-a-(-4-a-b) = 23
-2b = 10+4
-2b = 14 /:(-2)
b = -7
-----
4a-2(-7)-4-a-(-7) = 23
3a+14-4+7 = 23
3a = 23-17
3a = 6 /:3
a = 2
----
c = -4-a-b = -4-2-(-7) = -6+7 = 1
c = 1
f(x) = 2x²-7x+1
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f(x) = ax²+bx+c
f(1) = -4, f(-1) = 10, f(-2) = 23
-4 = a·1²+b·1+c
10 = a·(-1)²+b·(-1)+c
23 = a·(-2)²+b·(-2)+c
a+b+c = -4 => c = -4-a-b
a-b+c = 10
4a-2b+c = 23
a-b-4-a-b = 10
4a-2b-4-a-(-4-a-b) = 23
-2b = 10+4
-2b = 14 /:(-2)
b = -7
-----
4a-2(-7)-4-a-(-7) = 23
3a+14-4+7 = 23
3a = 23-17
3a = 6 /:3
a = 2
----
c = -4-a-b = -4-2-(-7) = -6+7 = 1
c = 1
----
f(x) = 2x²-7x+1
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