SOLUTION: Can somebody show me how to find the coordinates of the vertex of the graph of{{{ 2^2-6x+2 }}} ? Thank you!

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Can somebody show me how to find the coordinates of the vertex of the graph of{{{ 2^2-6x+2 }}} ? Thank you!      Log On


   



Question 6736: Can somebody show me how to find the coordinates of the vertex of the graph of+2%5E2-6x%2B2+ ?
Thank you!

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
There are several ways to find the coordinates of the vertex of a parabola.
I'm assuming that there is a typographical error in your question.
You probably meant to type:
y+=+x%5E2+-+6x+%2B+2
If not, your question is meaningless.
If this is the equation you intended, then read on.
1) You could graph the equation and find the location of the vertex of the parabola, or
2) You could put the equation into the form:
y+=+a%28x-h%29%5E2+%2B+k
and read the coordinates of the vertex from: (h, k)
This method requires that you perform an operation known as "completing the square".
3) You could put your equation into standard form ax%5E2+%2B+bx+%2B+c+=+0. Then find the x-coordinate of the vertex which is located at -b/2a.
Then you could get the y-coordinate by substituting the x-value into the equation y+=+x%5E2+-+6x+%2B+2 and solve for y.
Here's method 2.
Set you equation equal to zero.
x%5E2+-+6x+%2B+2+=+0 Subtract 2 from both sides.
x%5E2+-+6x+=+-2 Add the square of half the x-coefficient (-3)^2 = 9 to both sides.
x%5E2+-+6x+%2B+9+=+7 Factor the left side.
%28x+-+3%29%28x+-+3%29+=+7 Add 7 to both sides.
%28x+-+3%29%5E2+-+7+=+0 You now have the equation in the a(x-h)^2 + k form
%28x+-+h%29%5E2+%2B+k h = 3 and k = -7
The coordinates of the vertex are: (h, k) or (3, -7)