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.Com
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) (Show Source): You can put this solution on YOUR website!
If your quadratic equation is in standard form:< you can always find the x-coordinate of the parabola's vertex by:
In this problem with:, a = 1, b = -12, and c = 6, so...
Substituting a = 1 and b = -12
Simplifying, we get:
Now substitute this back into the given quadratic equation and solve for the y-coordinate.
Substitute x = 6.
Simplifying, we get:
The vertex is at (6, -30)
Let's check the graph for confirmation:
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