PERFECT SQUARES

Perfect squares determine the maxima or minima of your graph, If it is a minima, the "a" value will be positive and the parabola will open upwards. If it is a maxima, the "a" value will be negative and the parabola will open downwards. The method to convert the equation in standard form to a vertex form is called "Completing the squares". Here is how to solve the perfect squares:

Simple

First, you need to know when graphing a standard form equation, you need to evaluate into vertex form which is much more useful in graphing.

Y=x^2+8x+5

Standered Form

Y=(x+4)^2-11

Vertex Form

Y=(x^2+8x)+5

Second, you need to find out what number can make "x^2+8x" a perfect square. To find this, you need to divide the "h" value by 2 and square root it to find the answer to make the bracket expression a perfect square.

(8/2)^2=16

Y=(x^2+8x+16-16)+5

You after, add 16 into your equation but you can't add numbers into this expression. But what you can do is subtact 16 with 16. This will not affect or change anything in the equation. The negative "16" will be put into the equation for the next step.

Finally, your "16" is gone and your "-16" will be subtracted by 5 to become "-11". As we already know in the first step that our perfect square is "4". Your final answer should be y=(x+4)^2-11.

Y=(x+4)^2-11

Keep in mind , this is a perfect square so as we did in Unit 2 of "Completing the squares". your "^2" should be placed outside of the bracket.

Complex

WITH NEGATIVE

Steps will be completley the same but before you start "step 2" of the process, you would factor out the negative and the signs will switch. At the very end when you have your final answer, you will factor back in the negative only on the number you are evaluating for your previous "k" making your answer in vertex form. (Make sure your negative and modified "k" is always outside).

WITH "a" VALUE

Steps will be completley the same but before you start "step 2" you factor out the "a" value outside and the numbers must be divided by the coressponding "a" value. When you are at your final answer, Factor back in the "a" value to the number that you are evaluating to the previous "k" value and solve for "k" to get your final answer which should be in vertex form. (still keeping the "a" outside of the bracket and your modified "k" value outside as well).