document.write( "Question 313133: how do you get the x intercepts by evaluating the function in order to get other points to graph the function>??????
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document.write( "this is what I have so far:\r
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document.write( "f(x) =2x^2 +2
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document.write( " =2(x^2 +1) +2
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document.write( "f(x) = 2(x^2+1+.5-.5)+2
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document.write( "=2(x+2+1)+3(.5)+2
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document.write( "=2(x+3)^2 + 3.5\r
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document.write( "am i on the right track and if so what next in order to finish ?? \n" );
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Algebra.Com's Answer #223937 by Fombitz(32388)![]() ![]() You can put this solution on YOUR website! No, not on the right track. \n" ); document.write( "To get x-intercepts, set f(x)=y=0 and solve for x. \n" ); document.write( " \n" ); document.write( "Well there are no x such that when you square them and add 2 will give you zero. There are no x-intercepts for this parabola. \n" ); document.write( "It never crosses the x-axis. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "To get the y-intercept, set x=0 and solve for y. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "You can find the axis of symmetry and vertex by completing the square. \n" ); document.write( " \n" ); document.write( "Actually, it's already in vertex form, \n" ); document.write( " \n" ); document.write( "So the vertex is at (0,2) and it opens upwards. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "To get other points, just pick an x and calculate y. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( ".\r \n" ); document.write( "\n" ); document.write( " |