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

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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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
y=3x%5E2+%2B+15x+%2B+9
y=3%28x%5E2+%2B+5x%29+%2B+9
y=3%28x%5E2+%2B+5x%2Bb%5E2%29-3b%5E2+%2B+9
y=3%28x%5E2+%2B+5x%2Bb%5E2%29-3b%5E2+%2B+9-> compare %28x%5E2+%2B+5x%2Bb%5E2%29 to %28a%5E2+%2B+2ab%2Bb%5E2%29, you see that a=1,2ab=5; so, find b
2%2A1%2Ab=5->b=5%2F2
y=3%28x%5E2+%2B+5x%2B%285%2F2%29%5E2%29-3%285%2F2%29%5E2+%2B+9
y=3%28x%2B+5%2F2%29%5E2-3%2825%2F4%29+%2B+9
y=3%28x%2B+5%2F2%29%5E2-75%2F4+%2B+36%2F4
y=3%28x%2B+5%2F2%29%5E2-39%2F4
compare to y=a%28x-+h%29%5E2%2Bk where h and k are x and y coordinate of the vertex, and a%3E0 which means function has minimum
in your case h=-5%2F2 and k=-39%2F4
so, the vertex is at (-5%2F2,-39%2F4) or (-2.5,-9.75) and it is a minimum
+graph%28+600%2C+600%2C+-10%2C+10%2C+-12%2C+10%2C+3%28x%2B+5%2F2%29%5E2-39%2F4%29+


Answer by rothauserc(4718) About Me  (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
: