Yesterday’s post<\/a> opened the discussion about changing a quadratic function from standard form<\/p>\n y=ax2<\/sup> + bx + c<\/p>\n to vertex form<\/p>\n y=a(x-p)2<\/sup> + q<\/p>\n That post focused on the case where a=1:<\/p>\n y=x2<\/sup> +bx +c<\/p>\n Today, the case where a\u22601:<\/p>\n Example:<\/u><\/strong> Convert y=3x2<\/sup> + 24x – 5<\/u><\/strong> to vertex form.<\/p>\n Step 1:<\/strong> Like before, rewrite the equation with a space before the constant term:<\/p>\n y=3x2<\/sup>+ 24x -5<\/p>\n Step 2:<\/strong> Factor the coefficient of x2<\/sup> from both variable terms:<\/p>\n y=3(x2<\/sup> +8x ) -5<\/p>\n Step 3:<\/strong> Inside the brackets, complete the square (see my post here<\/a> for an introduction): take half the coefficient of the x term, square it, then add it inside the brackets.<\/p>\n In this case, the coefficient of x is 8. We take 4, square it to get 16, and then write that in the brackets.<\/p>\n y=3(x2<\/sup> + 8x + 16<\/span>) -5<\/p>\n Step 4:<\/strong> Very important<\/u>: Realize that when you added the number in the brackets, you really added that number times the number in front<\/em>. You need to subtract that product on the outside, to equalize.<\/p>\n In this case, we added 16 in the brackets; therefore, we really added 3(16)=48. To equalize, we must subtract 48 from the outside:<\/p>\n y=3(x2<\/sup> +8x +16<\/span>) -5 -48<\/span><\/p>\n Step 5:<\/strong> Now realize that the trinomial in the brackets is a perfect square. (Once again, see yesterday’s post.<\/a>)<\/p>\n y=3(x2<\/sup> +8x +16) – 53 becomes y=3(x+4)2<\/sup> -53<\/p>\n Apparently, y=3x2<\/sup> +24x -5, in vertex form, is y=3(x+4)2<\/sup> -53. The vertex is (-4,-53). (See my post here<\/a> about identifying the vertex.)<\/p>\n While this shows the general case of converting from standard form to vertex form, it’s an easy example. I’ll cover some more difficult ones in future posts.<\/p>\n HTH:)<\/p>\n