Example<\/strong><\/u>: Convert y=-5x2<\/sup> + 30x + 11 to vertex form.<\/p>\n Solution:<\/p>\n Step 1<\/strong>: As before, rewrite the equation with a space before the constant term.<\/p>\n y=-5x2<\/sup> + 30x +11<\/p>\n Step 2<\/strong>: Factor the coefficient of x2<\/sup> from both variable terms:<\/p>\n y=-5(x2<\/sup> -6x ) +11<\/p>\n Step 3<\/strong>: As before, we complete the square (see here<\/a> for details): we take half the coefficient of the x term, square it, then add that result inside the brackets.<\/p>\n In this case, the coefficient of x is -6. Half is -3, then (-3)2<\/sup> is 9.<\/p>\n y=-5(x2<\/sup> -6x +9<\/span>) +11<\/p>\n Step 4<\/strong>: Very important<\/u>: Adding the number inside the brackets actually means adding that number times the number in front<\/em>. To equalize, we must add the opposite to the outside.<\/p>\n In this case, adding 9 inside the bracket, I’ve really added -5(9)=-45<\/strong> to the equation. To equalize, I add 45<\/strong> to the outside.<\/p>\n y=-5(x2<\/sup> -6x +9<\/span>) +11 +45<\/span><\/p>\n Step 5<\/strong>: Realize that the three terms inside the brackets constitute a perfect square trinomial. (Once again, see this earlier post<\/a> for more detail.)<\/p>\n In this case,<\/p>\n y=-5(x2<\/sup> -6x +9) +11 +45 becomes y=-5(x-3)2<\/sup> +56<\/p>\n For a reader who might still be unsure about this process, here’s proof that it’s valid:<\/p>\n y=-5(x-3)2<\/sup>+56<\/p>\n (x-3)2<\/sup> = (x-3)(x-3) = x2<\/sup> -3x -3x +9 = x2<\/sup> -6x +9 (by the foil method<\/a>)<\/p>\n y=-5(x2<\/sup> -6x +9) + 56<\/p>\n Distribute the -5 into the brackets:<\/p>\n y=-5x2<\/sup> +30x -45 + 56<\/p>\n Now simplify:<\/p>\n y=-5x2<\/sup> +30x + 11<\/p>\n We arrive at the original equation, which means that our conversion of is valid.<\/p>\n Therefore, y=-5x2<\/sup> +30x + 11 becomes y=-5(x-3)2<\/sup> + 56 in vertex form. Its vertex is (3,56). (See my post here<\/a> about identifying the vertex.)<\/p>\n Converting y=ax2<\/sup> + bx + c to y=a(x-p)2<\/sup> + q can be counter-intuitive when a<0. In particular, step 4 above may be difficult to believe at first.\n\nIn a future post I'll likely visit another case of converting from standard form to vertex form. However, among this post and the previous two, the most important cases have been covered.\n\nHTH:)\n\nJack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":" The tutor gives another example of converting a quadratic function from standard to vertex form.\u00a0 Some people find the case a<0 to be tricky. My previous couple of posts (here and here) cover other examples of standard form to …<\/p>\n
\nMy previous couple of posts (here<\/a> and here<\/a>) cover other examples of standard form to vertex form. Today’s focus:<\/p>\n