SOLUTION: Find vertex of quadratic function. f(x)=x^2 - 12x+6 (x squared minus 12x plus 6) Here is what I have so far: step one: y=x^2 -12x+6 step two: y-6=x^2 - 12x Not sure whe

Algebra ->  Rational-functions -> SOLUTION: Find vertex of quadratic function. f(x)=x^2 - 12x+6 (x squared minus 12x plus 6) Here is what I have so far: step one: y=x^2 -12x+6 step two: y-6=x^2 - 12x Not sure whe      Log On


   

Question 148668: Find vertex of quadratic function. f(x)=x^2 - 12x+6
(x squared minus 12x plus 6)
Here is what I have so far:
step one: y=x^2 -12x+6
step two: y-6=x^2 - 12x
Not sure where to go after this!
Thanks!
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
If your quadratic equation is in standard form:y+=+ax%5E2%2Bbx%2Bc< you can always find the x-coordinate of the parabola's vertex by: x+=+%28-b%29%2F2a
In this problem with:y+=+x%5E2-12x%2B6, a = 1, b = -12, and c = 6, so...
x+=+%28-b%29%2F2a Substituting a = 1 and b = -12
x+=+%28-%28-12%29%29%2F2%281%29 Simplifying, we get:
x+=+6 Now substitute this back into the given quadratic equation and solve for the y-coordinate.
y+=+x%5E2-12x%2B6 Substitute x = 6.
y+=+%286%29%5E2-12%286%29%2B6 Simplifying, we get:
y+=+36-72%2B6
y+=+-30
The vertex is at (6, -30)
Let's check the graph for confirmation:
graph%28400%2C400%2C-5%2C15%2C-32%2C5%2Cx%5E2-12x%2B6%29