Steve had already answered this question for you.
http://www.jiskha.com/display.cgi?id=1363360907
What part of his answer did you not like ?
Determine without graphing, whether the given function has a maximum value and then find the value. f(x)=3x^2+30x-1
5 answers
I needed help learning how to find the value.
f(x) = 3(x^2 + 10x +25)-76
= 3(x+5)^2 -76
The function has no maximum. It has a minimum of -76 when x = -5.
= 3(x+5)^2 -76
The function has no maximum. It has a minimum of -76 when x = -5.
Hmmm. I explained that there is no maximum value. Don't expect to be able to find it.
Now, if you want to find the minimum value, try completing the square:
3x^2+30x-1
= 3(x^2 + 10x) - 1
= 3(x^2 + 10x + 25) - 1 - 75
That step is the key. We added 75 to complete the square, so we have to subtract it as well to avoid changing f(x)
= 3(x+5)^2 - 76
Now you can see that since (x+5)^2 is always at least zero (since squares are never negative), the smallest value for (x+5)^2 is when x = -5. In that case, f(x) = 3(0) - 76
So, f(-5) = -76, and it can never be less than that, because whatever x is, (x+5)^2 will be some positive number, making f(x) > -76
Now, if you want to find the minimum value, try completing the square:
3x^2+30x-1
= 3(x^2 + 10x) - 1
= 3(x^2 + 10x + 25) - 1 - 75
That step is the key. We added 75 to complete the square, so we have to subtract it as well to avoid changing f(x)
= 3(x+5)^2 - 76
Now you can see that since (x+5)^2 is always at least zero (since squares are never negative), the smallest value for (x+5)^2 is when x = -5. In that case, f(x) = 3(0) - 76
So, f(-5) = -76, and it can never be less than that, because whatever x is, (x+5)^2 will be some positive number, making f(x) > -76
-80, 5