document.write( "Question 143459: find the vertex and intercept coordinates of the following.\r
\n" ); document.write( "\n" ); document.write( "y=x^2-5x-4
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Algebra.Com's Answer #104410 by nabla(475)\"\" \"About 
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Set y=0 :\r
\n" ); document.write( "\n" ); document.write( "x^2-5x-4=0\r
\n" ); document.write( "\n" ); document.write( "Use the quadratic formula to find where y is zero (the x-intercepts):
\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"\r
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"1x%5E2%2B-5x%2B-4+=+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=%28-5%29%5E2-4%2A1%2A-4=41\".
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\n" ); document.write( " Discriminant d=41 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28--5%2B-sqrt%28+41+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+41+%29%29%2F2%5C1+=+5.70156211871642\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%28-5%29-sqrt%28+41+%29%29%2F2%5C1+=+-0.701562118716424\"
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\n" ); document.write( " Quadratic expression \"1x%5E2%2B-5x%2B-4\" can be factored:
\n" ); document.write( " \"1x%5E2%2B-5x%2B-4+=+1%28x-5.70156211871642%29%2A%28x--0.701562118716424%29\"
\n" ); document.write( " Again, the answer is: 5.70156211871642, -0.701562118716424.\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%2B-5%2Ax%2B-4+%29\"

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\n" ); document.write( "\n" ); document.write( "To find the y intercept we set x=0;\r
\n" ); document.write( "\n" ); document.write( "0^2-5(0)-4=y
\n" ); document.write( "It follows that y=-4. Thus, (0,-4) is the y-intercept.\r
\n" ); document.write( "\n" ); document.write( "The vertex is x=-b/(2a) when the equation is in form ax^2+bx+c=0. Thus, 5/2=x is the vertex. Now, find the corresponding y coordinate for x=5/2. \r
\n" ); document.write( "\n" ); document.write( "(5/2)^2-5(5/2)-4=25/4-50/4-16/4=-41/4
\n" ); document.write( "Thus, the vertex is at (5/2,-41/4). The graph is above.
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