SOLUTION: the vertex form equation of parabola identify coordinate v(h,k)
1)y=2x^2+32x+125
2)x^2-6x+y+9=0
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Question 398754: the vertex form equation of parabola identify coordinate v(h,k)
1)y=2x^2+32x+125
2)x^2-6x+y+9=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
the vertex form equation of parabola identify coordinate v(h,k)
1)y=2x^2+32x+125
2)x^2-6x+y+9=0
To identify the coordinates of a parabola, you can write the equation in this form: y=(x-h)^2+k, (h,k) being the (x,y) coordinates of the vertex. You must complete the square to do this.
(1)y=2x^2+32x+125
completing the square
factor out 2
y=2(x^2+16x+64)+125-128 (we added 2*64 to complete the square,so we must add -128 to even things out
y=2(x+8)^2-3
vertex: (-8,-3)
(2)x^2-6x+y+9=0
y=-(x^2-6x+ )-9
y=-(x^2-6x+9)-9+9 (we added -9 to complete the square,so we must add 9 to even
things out
y=-(x-3)^2)+0
vertex: (3,0)
Another method for finding the coordinates of the vertex is to use the formula:
x=-b/2a,a being the coefficient of the x^2-term, and b the coefficient of the x-term. In equation (1) above, x=-16/2=-8, then substitute this value for x in the equation to find y=128-256+125=-3. If you are just trying to find the coordinates of the vertex, this might be a more simple way, but the first method will yield more information about the parabola.You should learn both ways.
see the following graphs of the two parabola equations:
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