Hello, Pre Cal class. I'm the scribe for today. First off I want to say good job to all the scribes who have already posted!!.

Class today seemed a little short because all we did were the three questions on the board, the first two were:

1) f (x) = x2 - 6x - 3
2) f (x) = -2x2 + 8x +1

For each of the following
a) write in standard form (complete the square)
b) sketch the graph
c) Identify: i) vertex
ii) roots
iii) equation of the axis of symmetry
iv) domain
v) range
vi) direction of opening
vii) max/min value

After we had time to work on it our selves Mr.K asked us if we had any problems with these questions and some had problems finding the roots of question 1 and 2.

Question # 1

(* first we need the expression into standard form to help us find the vertex and the roots)

f (x) = x2 - 6x -3
(x) = x2 -6x - 3 +9 -9 -3
(x) = (x - 3)2 -12

Roots of # 1
0 = (x - 3)2 -12

12 = (x - 3)2
+-√12 = x - 3
3 +- √12 = x
3 +- 2√3 =x

So the roots in the parabola of question 1 is: x = 3 +- 2√3


a) f (x) = (x - 3)2 - 12
b) This is how the graph would look like:
<---- The rest of the chart is easy with the help of "Standard Form"




i) vertex = (3,-12)
ii) roots = x = 3 +- 2√3
iii) axis of symmetry = x=3
iv) domain = (-oo,oo)
v) range = [-12,oo)
vi) direction of opening = up
vii) max/min value = min y =-12





Question # 2

(*as we did in the first question we need to start by changing the expression into standard form to find the vertex and roots)

f (x) = -2x2 + 8x +1
(x) = -2 [x2 - 4x +4 -4 ] +1
(x) = -2 (x - 2)2 +8 +1
(x) = -2 ( x - 2)2 + 9

Roots of # 2
0 = -2 (x - 2)2 +9 2(x - 2)2 \ 2 = 9\2 (x -2)2 = 9\2 x-2 = +- √9\2
x = 2+- √9\2
x = 2+- 3\√2 (√2\√2)
x = 2+- 3√2\2

So the roots of the parabola for question # 2 is: x = 2+- 3√2\2


a) f (x) = -2 (x - 2)2 + 9
b) This is how the graph would look like:
<----

The rest of the chart is easy with the help of "Standard Form"



i) vertex = (2,9)
ii) roots = x = 2+- 3√2\2
iii) axis of symmetry = x=2
iv) domain = (-oo,oo)
v) range = (-oo,9]
vi) direction of opening = down
vii) max/min value = max y = 9

Most of us never did queston#3, and thought it had nothing to do with what we were learning for this unit.

3) Find the maximum rectanuglar area that can be enclosed by 160m of fencing if one side of the rectangle is an existing wall.

We found out it did have something to do with what we were learning. It has the same formulas we used to find what we found in the other two questions.

Homework for tonight is to do the worksheet we got "completing the square exercise" and to do exercise 7 #5 - 12

The next scribe is . . . . . . . . . Natnael. . . . .Have fun.!!