# 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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