# Week 9 – PreCalculus 11

This week in Pre Calc 11 we were reviewing for our Graphing unit test and our midterm exam. Some lessons that were taught in this week include  Equivalent forms of the quadratic function and  Modeling with quadratic functions. We also did our annual skills check my performance on the skills check was pretty good only missing 2 questions. The biggest thing that I struggled with this week was using a graph to form an equation and vice-versa. To solve my problem I had to practice and memorize converting from the 3 types of equations these include.

• $y= a (x-p)^2+q$ where (p,q) is the vertex coordinate
• $y= ax^2+bx+c$ where c represents the y-intercept
• $y= (x-x1)(x-x2)$ Where x1 and x2 represent the x-intercepts

Example: Step 1: Find Vertex Form

The first step we see how the vertex has transformed from (0,0). In this case, the p-value is -3 and the q value is -8. If we insert these numbers into the standard form we will be left with: $y= a(x+3)^2-8$

Now we just need to find the “a” value. According to the principle equation, the graph should go up by1 then by 3 and then by 5, etc. In this case, we see that the graph goes up by 2 and over one this means that the a value is congruent to $y=2x^2$

Step 1: Convert To General Form

Convert it from vertex form to general form. It is quite simple all you have to do is expand your equation and simplify again. Below you will see the process. $y= 2(x+3)^2 -8$ $y=2(x+3)(x+3)-8$ $y=2(x^2+6x+9)-8$ $y= 2x^2 +12x + 10$

Now you have the General Form of the equation which gives you your y-intercept

Step 3: Find Factored Form

The reason to use factored for is to find x-intercepts to verify the position of your graph. Its really quite simple all one has to do is factor the equation in from the general form.

Take out the 2 from the equation: $y=2x^2 +12x +10$ $y= 2(x^2 +6 +5)$

Now you must factor what is in the prackets to create two sets of brackets like so: $y=2(x+3)(x+2)$