document.write( "Question 156920: Graph the quadratic equation, label the ordered pairs for the vertex and the y-intercept\r
\n" ); document.write( "\n" ); document.write( "y= x^2 + x-2
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Algebra.Com's Answer #115694 by Alan3354(69443)\"\" \"About 
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Graph the quadratic equation, label the ordered pairs for the vertex and the y-intercept
\n" ); document.write( "y= x^2 + x-2
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"1x%5E2%2B1x%2B-2+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%281%29%5E2-4%2A1%2A-2=9\".
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\n" ); document.write( " Discriminant d=9 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-1%2B-sqrt%28+9+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%281%29%2Bsqrt%28+9+%29%29%2F2%5C1+=+1\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%281%29-sqrt%28+9+%29%29%2F2%5C1+=+-2\"
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\n" ); document.write( " Quadratic expression \"1x%5E2%2B1x%2B-2\" can be factored:
\n" ); document.write( " \"1x%5E2%2B1x%2B-2+=+%28x-1%29%2A%28x--2%29\"
\n" ); document.write( " Again, the answer is: 1, -2.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-2+%29\"

\n" ); document.write( "\n" ); document.write( "The vertex is at the minimum. To solve for the minimum, set the 1st derivative to 0.
\n" ); document.write( "2x+1 = 0
\n" ); document.write( "x = -1/2
\n" ); document.write( "Sub -1/2 for x into the eqn.
\n" ); document.write( "y(min) = (-1/2)^2 +(-1/2) -2
\n" ); document.write( "y(min) = 1/4 -1/2 - 2
\n" ); document.write( "= -2 1/4 = -9/4
\n" ); document.write( "So the vertex is (-1/2,-9/4)
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\n" ); document.write( "The y-intercept is easier, it's where x = 0. Sub 0 for x in the original eqn:
\n" ); document.write( "y = 0 + 0 -2 = -2
\n" ); document.write( "So the point is (0,-2)\r
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