Quadratic functions are often written in standard form<\/p>\n
y=ax2<\/sup>+bx+c<\/p>\n but more useful is vertex form<\/p>\n y=a(x-p)2<\/sup>+q<\/p>\n A quadratic function in standard form can be converted to vertex form. One method is completing the square; see my post here<\/a> for an introduction.<\/p>\n Example<\/span><\/strong>.<\/p>\n Convert y=x2<\/sup> -10x -1 to vertex form.<\/p>\n Step 1<\/strong>: Rewrite the equation, leaving a space before the constant:<\/p>\n y=x2<\/sup> -10x\u00a0\u00a0\u00a0\u00a0\u00a0-1<\/p>\n Step 2<\/strong>: Take half the coefficient of the x<\/strong> term and square it.<\/p>\n Here, the coefficient of the x<\/strong> term is -10. Half is -5; (-5)2<\/sup>=25.<\/p>\n Step 3<\/strong>: Both add and subtract the squared value to the right side of the equation:<\/p>\n y=x2<\/sup> -10x\u00a0 +25<\/span>\u00a0-1 -25<\/span><\/p>\n Step 4<\/strong>: Notice that the first three terms on the right side form a perfect square trinomial:<\/p>\n y=(<\/span>x2<\/sup>\u00a0-10x\u00a0+25<\/span>)<\/span>\u00a0-1\u00a0-25<\/span> \u00a0becomes y=(x-5)2<\/sup> \u00a0\u00a0-26<\/p>\n [Reason: (x-5)2<\/sup>=(x-5)(x-5) = x2<\/sup> -5x -5x +25 = x2<\/sup> -10x +25, by foil method<\/a>]<\/p>\n Step 5<\/strong>: Apparently, vertex form of y=x2<\/sup> -10x -1 is y=(x-5)2<\/sup> -26. The vertex is (5,-26). (To identify the vertex, see my post here<\/a>.)<\/p>\n Next post, I’ll be covering the case where the coefficient of x2<\/sup> is other than 1. It will be dependent on the technique described here.<\/p>\n HTH:)<\/p>\n