document.write( "Question 19436: Im not sure how to do this problem \r
\n" ); document.write( "\n" ); document.write( "Graph: $\"y=x%5E2-4x%2B1\"$ \n" ); document.write( "

Algebra.Com's Answer #9395 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
You can find the x-coordinate of the vertex of this parabola from: $\"x+=+-b%2F2a\"$
\n" ); document.write( " $\"x+=+-%28-4%29%2F2\"$ = 2
\n" ); document.write( "The y-coordinate is found by substituting this value of x (2) into the quadratic equation and solving for y:
\n" ); document.write( " $\"y+=+2%5E2+-+4%282%29+%2B+1\"$
\n" ); document.write( " $\"y+=+4+-+8+%2B+1\"$
\n" ); document.write( " $\"y+=+-3\"$ \r
\n" ); document.write( "\n" ); document.write( "The vertex is at (2, -3)\r
\n" ); document.write( "\n" ); document.write( "Then you could find the x-intercepts (there will be two of 'em) by letting y = 0 and solving for the roots.\r
\n" ); document.write( "\n" ); document.write( " $\"x%5E2+-+4x+%2B+1+=+0\"$ Use the quadratic formula to solve: $\"x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"x+=+%28-%28-4%29%2B-sqrt%28%28-4%29%5E2+-+4%281%29%281%29%29%29%2F2%281%29\"$
\n" ); document.write( " $\"x+=+%284%2B-sqrt%2812%29%29%2F2\"$
\n" ); document.write( " $\"x+=+2%2Bsqrt%283%29\"$ and $\"x+=+2-sqrt%283%29\"$ \r
\n" ); document.write( "\n" ); document.write( "Here's what the graph will look like:
\n" ); document.write( " $\"graph%28300%2C200%2C-10%2C10%2C-10%2C10%2Cx%5E2-4x%2B1%29\"$ \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );