Answer:
[tex]f(x)=-2x^2-3x-9[/tex]
Step-by-step explanation:
Let [tex]f(x)=ax^2+bx+c[/tex] be the quadratic function.
[tex]f(-2)=-11[/tex] implies that [tex]a(-2)^2+b(-2)+c=-11[/tex]
[tex]4a-2b+c=-11......(1)[/tex]
Similarly, [tex]f(1)=-5\implies a(1)^2+b(1)+c=-5\implies a+b=-5....(2)[/tex]
and [tex]f(2)=-23\implies a(2)^2+b(2)+c=-23\implies 4a+2b+c=-23...(3)[/tex]
From equation 2, [tex]a=-b-5...(4)[/tex]
Put (4) into (1) to get:
[tex]4(-b-5)-2b+c=-11\implies -6b+c=9....(5)[/tex]
Put (4) in (3) to get:
[tex]4(-b-5)+2b+c=-23\implies -2b+c=-3---(6)[/tex]
Subtract (6) from (5) to get:
[tex]-4b=12[/tex]
[tex]b=-3[/tex]
Put b=-3 in (4) to get:
a=--3-5=-2
Put b=-3 in to (6) to get:
-2(-3)+c=-3
6+c=-3
c=-6+-3=-9
Therefore the required equation is [tex]f(x)=-2x^2-3x-9[/tex]
