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solving equations algebraically

kevinryans1

New member
Joined
Feb 1, 2010
Messages
6
I have a problem with the question:

Find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x - 8x + 3.

I do not understand how to solve it algebraically. It would be great if you can help. Please and Thank You.
 

arthur ohlsten

Full Member
Joined
Feb 20, 2005
Messages
854
the equation as written is not a parabola. I assume you meant:
y=-2x^2-8x+3
we want the equation in the form
y=p[x-a]^2+b parabola open up if p is + open down if p=- vertex at a,b

y=-2x^2-8x+3 factor out -2 from first two terms
y=-2[x^2+4x] +3 complete the square
y=-2[x^2+4x+4]+8+3
y=-2[x+2]^2 + 11 answer

Arthur
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,091
kevinryans1 said:
I have a problem with the question:

Find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x - 8x + 3.

I do not understand how to solve it algebraically. It would be great if you can help. Please and Thank You.
The axis of symmetry is the vertical line through the vertex.

From the equation above, provided by Arthur, you can find the vertex and the axis of symmetry.

Let us know if you are still stuck.
 
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