document.write( "Question 29731: I need help graphing these functions could I get some help please?\r
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\n" ); document.write( "\n" ); document.write( "y=4x^2\r
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\n" ); document.write( "\n" ); document.write( "and graph function y=x^2-2 \n" ); document.write( "
Algebra.Com's Answer #16552 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
For the first function: \"y+=+4x%5E2\", you know (or should know) that this is the equation for a parabola that opens upwards (the coefficient of x^2 is positive) and the x-coordinate of the vertex is given by: \"x+=+-b%2F2a%29\".
\n" ); document.write( "The a and b come from the standard form for the quadratic equation: \"ax%5E2+%2B+bx+%2B+c\", so the vertex of the parabola is: \"x+=+0\" because b = 0, and, in your equation, when x = 0 then y = 0, so the vertex of the parabola is at (0, 0), the origin of the coordinate system.
\n" ); document.write( "The x-intercepts can be found by setting y = 0 and solving for x.
\n" ); document.write( "\"0+=+4x%5E2\"
\n" ); document.write( "\"x+=+0\" and \"x+=+0\" or, there are no x-intercepts which makes sense when you realise that if the parabola opens upwards and the vertex is at the origin, then the parabola never crosses the x-axis.
\n" ); document.write( "Here's the graph:
\n" ); document.write( "\"graph%28300%2C200%2C-3%2C3%2C-3%2C5%2C4x%5E2%29\"\r
\n" ); document.write( "\n" ); document.write( "For the second function, you can apply a similar analysis and find that the parabola opens upwards but the vertex is at (0, -2) and the x-intercepts are at x = \"sqrt%282%29\" and x = -\"sqrt%282%29\"
\n" ); document.write( "Here's the graph:
\n" ); document.write( "\"graph%28300%2C200%2C-5%2C5%2C-5%2C5%2Cx%5E2-2%29\"
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