document.write( "Question 390961: Graph the parabola y=2x^2+4x-6 on graph paper \n" ); document.write( "

Algebra.Com's Answer #277279 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "y=2x^2+4x-6  |determining the x-intercepts (when y=0)
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document.write( " 2x^2+4x-6 = 0
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document.write( "factoring
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document.write( "(2x -2)(x+3)=0 Note:SUM of the inner product(-2x) and the outer product(6x) = 4x
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document.write( " (2x -2)=0  x = 1  x-intercept at Pt(1,0)
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document.write( " (x-3)=0    x= -3    x-intercept at Pt(-3,0)\r
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document.write( "Note: the vertex form of a parabola,  where(h,k) is the vertex
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document.write( "y=2x^2+4x-6   |completing the square to put into vertex form
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document.write( "y = 2(x^2+2)-6
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document.write( "y= 2[(x+1)^2 -1] -6
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document.write( "y= 2(x+1)^2 -8   Vertex is at Pt(-1,-8)  parabola opens upward (2>0)
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