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Find the largest value of c such that 1 is in the range of f(x)=x^2-5x+c.

xXxTenTacion  Aug 24, 2018
 #1
avatar+91019 
+1

Since this parabola turns upward....1 will still be in the range when, at a minimum,  (m , 1)  is the

vertex

 

We have the form   ax^2 + bx + c

 

To  find "m"  we   can use    -b/ [2a]  =       5 / (2 * 1)   =  5/2  = m

 

So  (5/2, 1)  is the vertex

 

So....subbing this into the function for x  and f(x)  to find " c,"  we have

 

1  = (5/2)^2  - 5(5/2) + c

1 = 25/4 - 25/2 + c

1  = 25/4 - 50/4 + c

1  = -25/4  + c

1 + 25/4  = c

29/4   =  c

 

Here's a graph to show this : https://www.desmos.com/calculator/gbhwg6nerl

 Check for yourself that if c > 29/4,  then  1 will be out of range

 

cool cool cool

CPhill  Aug 24, 2018
edited by CPhill  Aug 24, 2018

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