SOLUTION: y=3x^2 + 15x + 9 Find the vertex and if it is maximum or minimum Do it by finding the square please!

Question 1048459: y=3x^2 + 15x + 9
Find the vertex and if it is maximum or minimum
Do it by finding the square please!
Found 3 solutions by ewatrrr, MathLover1, rothauserc:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
y=3x^2 + 15x + 9
y=3(x^2 + 5x) + 9
y=3(x+ (5/2))^2 - 75/4 + 36/4
y = 3(x+ (5/2))^2 - 39/4
V(-2.5, - 9.75)

Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

-> compare to , you see that ,; so, find
->

compare to where and are and coordinate of the vertex, and which means function has minimum
in your case and
so, the vertex is at (,) or (,) and it is a minimum

Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
y = 3x^2 + 15x + 9
:
y - 9 = 3x^2 + 15x
:
y - 9 = 3(x^2 +5x)
:
y - 9 + 3(25/4) = 3(x^2 +5x + 25/4)
:
y - 36/4 + 75/4 = 3(x + 5/2)^2
:
y + 39/4 = 3(x + 5/2)^2
:
y = 3(x + 5/2)^2 - 39/4
:
y = 3( x - (-5/2))^2 - 39/4
:
*********************************************
vertex is at (-5/2, -39/4) which is a minimum
*********************************************
:
we can check this by using
:
x = -b/2a = -15 / 6 = -5/2
:
y = 3(-5/2)^2 + 15(-5/2) + 9
:
y = 75/4 -150/4 + 36/4 = -39/4
:
our answer (-5/2, -39/4) checks
:

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