Answer to Question #4954 in Calculus for sean
Question #4954
the graph of a function is on the interval [-3,3] and f(x)= x^2 +ax+b the absolute minimum is (-2,2) find f(1)
Expert's answer
-b/2c=-2
c=1 then -b=-4 then
b=4
4+2a+b=2
2a=-6
a=-3
x^2-3x+4
f(1)=1-3+4=2
c=1 then -b=-4 then
b=4
4+2a+b=2
2a=-6
a=-3
x^2-3x+4
f(1)=1-3+4=2
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