SOLUTION: Find {{{a}}} such that {{{f(x)=ax^2+5x+6}}} has a minimum value of {{{169/24}}}

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Question 889068: Find such that has a minimum value of
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
Complete the Square:



, almost finished standard form.

The extreme value happens when and ; there you can solve for a.




, LCD is 16*3a.






.
Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!
Find such that has a minimum value of 


(h, k) is the coordinate point of the vertex of the parabola

h , or x-coordinate of vertex = , or 



, with 

k, or y-coordinate of vertex = 



    



 ------ Multiplying right-side by LCD, 4a

   



Since , and minimum value = , then: 

169(4a)  = 24(- 25 + 24a) ------ Cross-multiplying

676a = - 600 + 576a

676a – 576a = - 600

100a = - 600

, or 

Since a = - 6, then this equation DOES NOT DEPICT a minimum, but a maximum instead, as a < 0.



You can do the check!! 



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