Find the min/max of f(x)=5x^2+30x+49
find f'(x) and equate it to 0 to find the critical points. Then find f''(x) and then find the value of f''(x) at these points if the value is >0 then the point is minima and if the value is <0 then the point is maxima
This is not for calculus this is for algerbra we don't do derivatives...
Then I m sorry because I know only this method
actually you are only looking for the vertex
Critical points is the best method. But you're looking for the vertex. Can you factor this function? ;-)
|dw:1340199943871:dw| you only have a min value
Darn it precal... -_- lol
|dw:1340199980619:dw|
|dw:1340200003010:dw|
now find f(-3) and that is your y value of your vertex
On sidenote, if you have a quadratic function \[f(x) = ax^{2} + bx + c\] Where a is not 0, then if a>0, you will only have a minimum value for y if a<0, you will only have a maximum value for y
instructions were probably given (as general) to a group of problems, teacher is probably trying to determine if student understands the difference between a min or max in quadratics
I understand, but good to know those, anyway, right? :)
yes, in calculus max and min can be used as local max or local min or absolute max or absolute min
we use that too in precal @terenzreingz
But thanks for your help everyone! I really do appreciate it :)
anytime :)
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