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Find the largest value of c such that -2 is in the range of \(f(x)=x^2+3x+c\) .

 Aug 27, 2018
 #1
avatar+128473 
+3

OK...similar to the last one

 

This parabola also turns upward....

 

The x coordinate of the vertex is    -3/2

 

So...put this into the function  and set up an inequality where the function is  greater than  -2

Thus...when we find solutions....we wil exclude those from the possible answers...so we have

 

(-3/2)^2 + 3(-3/2)  + c  > -2    simplify

 

9/4 - 9/2 + c > -2

 

-9/4 + 2 > - c

 

-9/4 + 8/4 > - c

 

-1/4 > - c    divide both sides by -1 and reverse the inequality sign

 

1/4 <  c

 

So...-2  will be in range whenever c  ≤ 1/4

 

See the graph, here : https://www.desmos.com/calculator/8uqpiabiz6

 

Note that when c = 1/4....-2 is still in the range

But..for instance..when c  = 1/2, -2 is out of range

 

 

cool cool cool   

 Aug 27, 2018
edited by CPhill  Aug 27, 2018
 #2
avatar+816 
+3

Thanks! I understand better!

mathtoo  Aug 27, 2018

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