SOLUTION: How do I solve a problem like this: find vertex of parabola y=3x^2+6x-12 Thanks, J

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Question 468414: How do I solve a problem like this:
find vertex of parabola
y=3x^2+6x-12
Thanks,
J

Found 2 solutions by MathLover1, josmiceli:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
y=3x%5E2%2B6x-12......standard form
When a function is in standard form rather than vertex form , we cannot simply look at the function and find the vertex of (h, k). There are two ways to approach this problem.
Using a formula: The x-value of the vertex can be found using the formula h=-b%2F2a y-value.
y=3x%5E2%2B6x-12......a=3, b=6 and c=-12
h=-b%2F2a
h=-6%2F2%2A3
h=-6%2F6
h=-1
y=3x%5E2%2B6x-12
y=3%28-1%29%5E2%2B6%28-1%29-12
y=3%281%29-6-12
y=3-18
y=-15
the vertex is (h, k)=(-1, -15)
+graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+10%2C+3x%5E2%2B6x-12%29+





Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+y+=+3x%5E2+%2B+6x+-+12+
The vertex is at either the minimum or maximum of the
parabola. The coefficient of the squared term is positive,
so it will have a minimum.
When the equation is in the form +y+=+ax%5E2+%2B+bx+%2B+c+.
The x-coordinate of the minimum is at +-b%2F%282a%29+
+a+=+3+
+b+=+6+
x = +-6+%2F+%282%2A3%29+
x = +-1+
Now to find the y-coordinate,
+y+=+3x%5E2+%2B+6x+-+12+
+y+=+3%2A%28-1%29%5E2+%2B+6%2A%28-1%29+-+12+
+y+=+3+-+6+-+12+
+y+=+-15+
The vertex is at (-1,-15)
Here is a plot:
+graph%28+400%2C+400%2C+-5%2C+5%2C+-16%2C+5%2C+3x%5E2+%2B+6x+-+12%29+