SOLUTION: Find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2-8x+3
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Question 457112: Find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2-8x+3 Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2-8x+3
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y=-2x^2-8x+3
completing the square
y=-2(x^2+4x+4)+3+8
y=-2(x+2)^2+11
Standard form of parabola: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. A is a multiplier which affects the steepness or narrowness of the parabola.
For given parabola, coordinates of the vertex at (-2,11), axis of symmetry, x=-2
Minus sign means parabola opens downward.
see graph below as a visual check on answer
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