document.write( "Question 201323: Find the x-intercepts and y-intercepts of the graph of the equation.\r
\n" ); document.write( "\n" ); document.write( "y=x^2+6
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Algebra.Com's Answer #151536 by jsmallt9(3758)\"\" \"About 
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X-intercepts are the point or points, if any, where a graph crosses the x-axis. If one thinks about the coordinates of the points on the x-axis you will find that while the x-coordinates can be any number, the y-coordinates of all the points on the x-axis are zero. So to find an x-intercept, substitute zero for the y, NOT THE X, in the equation. Your equation, y=x^2+6, becomes:
\n" ); document.write( "(0) = x^2 + 6
\n" ); document.write( "Solving for x:
\n" ); document.write( "Subtract 6
\n" ); document.write( "0 - 6 = x^2 + 6 -6
\n" ); document.write( "-6 = x^2
\n" ); document.write( "Since it is impossible to square a Real number and get a negative number like -6, this equation has no solution. This means that there are NO x-intercepts for your equation, y=x^2+6.
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\n" ); document.write( "Y-intercepts are where a graph crosses the y-axis. Using reasoning similar to that used for x-intercepts we find that all points on the y-axis have an x-coordinate which is zero. So to find a y-intercept, substitute a zero for the x, NOT THE Y, in the equation. Your equation, y = x^2 + 6, becomes:
\n" ); document.write( "y = (0)^2 + 6
\n" ); document.write( "y = 0 + 6
\n" ); document.write( "y = 6
\n" ); document.write( "So the only y-intercept is (0,6).
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