Sup

Hi guys, it's A.J. from Mr.P's 30s pre-cal class. Today in class we finished the last page of our unit one booklet and started are unit one review. Wow all we did was a review so what do I talk about. Kay well I'll review Transformations of Quadratic functions 3 and Applications example 7.

"During the summer months Terry makes and sells necklaces on the beach. Last summer he sold necklaces for $10 each. His sales averaged 20 per day. Considering a price increase, he took a small survey and found that for every dollar increase he would lose two sales per day. If the material for each necklace costs $6, what should the selling price be to maximize profit."

Step one
find the equation and let one value equal "x"
20 = necklaces
2 = lost from price increase
x = number of the one dollar increase
10 = how much the necklaces are
6 = money for the material
THUS
(20-2x)(10-6+x)
or
(20-2x)(4+x)

Step 2
First
Outside
Inside
Last

Thus getting
80+20x-8x-2x²

Then change it around
-2x²+20x-8x+80
Then combine the alike terms
-2x²+12x+80
That's out equation

Step 3
Solve for X





















Step 4
Place X in the equation

-2(3)²+12(3)+80
3²= 9
9 . 2- = -18

12 . 3 = 36

-18+36+80=98

Step 5
Add the sentence
Terry made a maximum profit of $98

yeah. so I hope this helps guys sorry I just came home from watching volleyball. Oh yes don't forget to vote for the people in September. I have alot of grammar mistakes. I'm going to say "no don't read it" tomorrow