Question 439903
Graphing Parabolas, for example this y= -x^2-2x+1? 
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If you don't know much about quadratics and parabolas
the best you can do is plot points.
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If you know somethings you could find the y-intercept (0,1);
you could find the x-intercepts by solving -x^2-2x+1 = 0;
you could find the vertex at (-b/(2a),f(-b/(2a));
you could notice the parabola opens downward because the
coefficient of x^2 is negative.
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{{{graph(400,300,-10,10,-10,10,-x^2-2x+1)}}}
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Cheers,
Stan H.