SOLUTION: I submitted the following problems for some help. I got no response and wanted to try again. If someone could help me I would be thankful.find the axis of symmetry and the vertex f

Algebra ->  Rational-functions -> SOLUTION: I submitted the following problems for some help. I got no response and wanted to try again. If someone could help me I would be thankful.find the axis of symmetry and the vertex f      Log On


   

Question 160762: I submitted the following problems for some help. I got no response and wanted to try again. If someone could help me I would be thankful.find the axis of symmetry and the vertex from each parabolic function. A. y=2x^2 -24x+22 B.y=x^2-3x-5
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
read this --> http://www.regentsprep.org/Regents/Math/conics/LPara.htm
Parabolas are of the form: y = ax^2 + bx + c
If a is positive, the parabola opens upward and has a minimum point.
The axis of symmetry is
x = (-b)/2a

If a is negative, the parabola opens downward and has a maximum point.
The axis of symmetry is
x = (-b)/2a.
Once you have the axis, use that value of x to find the vertex.

1) +y=2x%5E2+-24x%2B22+
Since a is positive, use x+=+%28-b%29%2F%282a%29
b is -24, so -b is 24. a is 2
x+=+24%2F%282%2A2%29
x+=+24%2F4
+x+=+6
Now solve for y when x=6
y+=+2%2A6%5E2+-+24%2A6+%2B+22
y+=+72+-+144+%2B+22
y+=+-50
So the vertex is (6,-50)
graph%28400%2C400%2C+-10%2C+10%2C+-100%2C+100%2C+2x%5E2+-24x%2B22%29
You can do the second one using the same method