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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jul 2007 22:00:18 IST
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How to draw the graph of polynomial like 2x^2 + 12x +16 ??? i know it by the factorization method and coverting the equation in whole square form by adding and subtracting something...... i want to know an alternative method and also shortways for solving any graph? ALL ACCEPTABLE ANSWERS WILL BE SALUTED FOR SURE.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jul 2007 22:20:53 IST
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Find the derivative of the polynomial w.r.t x. Equate the derivative to 0. The value of x obtained, when being put in in the polynomial will give its max.(A<0) or min. value(A>0). After this, the graph can be drawn easily. For ex; y=2x^2+12x+16 Derivative=4x+12=0 which gives x=-3 On putting x=-3 in y=2x^2+12x+16, we get min. value of this polynomial. which comes out to be -2. Plzzz rate me if this was helpful.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jul 2007 22:33:00 IST
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Arey!! how can i know the shape of the graph by knowing the min. point plzz explain a little more.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jul 2007 07:42:43 IST
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Plzz tell me if there is any other method other than putting the values in hte equation and drawing the graph???
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jul 2007 10:04:15 IST
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I think the the method which u use is the most easy and the appropriate one. Convert into perfect square.Stick to that itself.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jul 2007 16:04:32 IST
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Then, the simplest and fastest method of doing this is as follows: Find the co-ordinates (-b/2a,-d/4a) of the polynomial ax^2+bx+c, where D is the Discriminant. This co-ordinate will be the vertex of the parabola of the polnomial, and the rest of the graph will be symmetric about this point. The graph could be drawn by taking one pt. ahead(x-coordinate) of the point (-b/2a,-d/4a) and find the y coordinate, and one point before. Drawing a smooth curve will give the graph. Plzzzz rate me if this was useful.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jul 2007 21:03:39 IST
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1) see the symmetry of function. here its neither even nor odd ...so no symmetry(abt axes)
2)find y when x=0 and find x when y=0 here , y=0 gives x=-2,-4
3) find the regions where y>0 and where y<0
here putting 2(x+2)(x+4)>0 gives x=(-inf,-4)+(-2,inf)...for this region y lies above x axis for other x,y lie below x axis
4) find limit x approaching infinity and -infinity.
here both are infinity and -infinity resp.
5) u may find points of local minima or maxima if u want...
now u can plot the graph....using the fact that all polynomial functions are continuous in their domain...
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Jul 2007 01:01:45 IST
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The curve is a second degree equation in x which tells us it is a parabola Find the minima And hence find the co-ordinates of the vertex of the parabola...... Now the equation of the axis would be x= abcissa of the vertex The graph is now symmetric about the vertex and because the coefficient of x^2 is +ve, it would be upper parabola If it were negative, it would have been upside down, rest all remainin the same........
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HOPE U GOT IT... |
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