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avatar+814 

For what values of b is -2 not in the range of the function \(f(x)=x^2+bx+2\)? Express your answer in interval notation.

 Aug 26, 2018
 #1
avatar+101796 
+3

x^2 + bx + 2

 

This parabola turns upward....so...the x coordinate of the vertex is given by

 

-b/ 2  = x

 

So....we'd ike to solve this

 

(-b/2)^2 + b(-b/2)  + 2  >  - 2    simplify

 

b^2/4 - b^2/2 + 4  >  0     

 

-b^2/4 + 4 > 0

 

-b^2 + 16 > 0  multiply through by  -1  and change the inequality sign

 

b^2 - 16 < 0

(b - 4) ( b + 4) < 0

The values  that make this true  are  ( -4, 4)

 

So...the values that put  -2 out of range  are  (-4, 4)

 

 

cool cool cool

 Aug 26, 2018
edited by CPhill  Aug 26, 2018
 #2
avatar+814 
+2

Yes! Correct!

 Aug 27, 2018

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