Precal Lesson

In today's class, we started off class by correcting some of the questions in Assignment #5. Luckily, questions that contained equations such as:

"The sum of the digits of a certain two-digit number is 12. Reversing its digits decreases the number by 54. What is the number?"

ARE OMITED. However, this assignment is postponed to December 20, 2010 (tomorrow).

We continued class learning about a new lesson called: Quadratic-Linear System. A Quadratic-Linear system has one "linear" and "quadratic" equation and its solutions can be graphed or can be solved algebraically.




1. There could be 1 solution (they intersect in exactly one spot)



















2. There could be 2 solutions (they intersect in exactly 2 spots)



















3. There could be no solutions (they do not intersect)


















Note: The quadratic functions do not always have to be parabolas, they include Circles, Elipses and Hyperbolas


Example 1: Solve the following system of equations graphically and algebraically.

y=x-1
y=x^2-6x+9

-ALGEBRAICALLY-

You can solve the following system of equations algebraically ! Here's how :


Step 1: Number these equations like you would do when solving systems of equations


y=x-1 [1]
y=x^2-6x+9 [2]


Step 2: Substitute the first equation into the second one, Since it already gives you the "y-variable" by itself


x-1=x^2-6x+9


Step 3: Combine like terms and equade the equation to 0 (in this case)


x^2-7x+10=0


Step 4: Simplify the Trinomial and then find the x-intercept(s)

x^2-7x+10=0
(x-5)(x-2)=0

x-5=0 or x-2=0
x=5 x=2


Step 5: Substitute the "x" variables into the linear formula to find the y-intercept(s)


y=5-1 y=2-1
y=4 y=1


The solutions are (5,4) and (2,1)


That is all for today, Remember to finish assignment #5 and hope you all have a wonderful Christmas Holiday ! OH, ..dont forget the project.