document.write( "Question 466195: graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflection). \r
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document.write( "[Hint: If necessary, write f in the form f(x) = a(x – h)2 + k.]\r
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document.write( "f(x) = –2x2 + 6x + 2 \n" );
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Algebra.Com's Answer #319526 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflection). \n" ); document.write( "[Hint: If necessary, write f in the form f(x) = a(x – h)2 + k.] \n" ); document.write( "f(x) = –2x2 + 6x + 2 \n" ); document.write( ".. \n" ); document.write( " –2x^2 + 6x + 2 \n" ); document.write( "completing the square \n" ); document.write( "-2(x^2-3x+9/4)+2+9/2 \n" ); document.write( "-2(x-3/2)^2+13/2 \n" ); document.write( "This is a parabola of the standard form, y=A(x-h)^2+k, with h=3/2 and k=13/2, and it opens downward. \n" ); document.write( "see the graph of this parabola as follows: \n" ); document.write( ".. \n" ); document.write( " \n" ); document.write( ".. \n" ); document.write( "The next graph shows the result of removing the coefficient A=2. Notice how the curve is a little wider or less steep. The larger this coefficient, the steeper the curve. \n" ); document.write( ".. \n" ); document.write( " \n" ); document.write( "\n" ); document.write( ".. \n" ); document.write( "The next graph shows the result of removing (h,k) the (x,y) coordinates of the vertex. \n" ); document.write( ".. \n" ); document.write( " \n" ); document.write( ".. \n" ); document.write( "Removing the negative coefficient \n" ); document.write( ".. \n" ); document.write( " \n" ); document.write( ".. \n" ); document.write( " |