Grade 11 Pre-Cal E

Quadratic Functions [Cont'd]

Greetings, I, Mr.Lawrence, am today's scribe.

Review; In general, y = ax² or x = ay² [Negative 'a' value will open the x = ay² function to the left.]

So, If a > 0 parabola opens upward ( x² )
If a > 0 parabola opens right ( y² )


Therefore, If a < 0 parabola opens downward ( x² )
If a < 0 parabola opens left (y²)


You should recognize a > 1 ; the parabola will be narrower ( skinnier )

*Note: a represents an absolute value


a < 1 ; the parabola will be wider.
The following parabolas are:
y = x² ; Vertex (0,0) AOS ; X = 0 Domain ( -00, 00) Range [0,00)


y = x² + 3 ; Vertex (0,3) AOS ; X = 0 Domain (-00,00) Range [3,00]


y = x² - 5 ; Vertex (0,-5) AOS ; X = 0 Domain (-00,00) Range [-5,00)


As we can see, there is a relationship.


Equation in the form of y = ax² + k
- The K Value directs/shifts the graph vertically.
-> If "K" is positive, Graph goes up
-> If "K" is negative, Graph goes down


AOS of all graphs is X = 0

The vertex is (0, K) so we know right away, what the vertex is just by looking at the equation.


Now the next equation is y = (x+1) ² and y = (x-1)²


This equation is in the form of y = (x-h) ² "h" is affecting the graph, shifting it horizontally.

*Note:
1) Axis of semmetry is the line x = h
2)Coordinates of vertex are (h,0)
3)If "h"> 0 parabola shifts to the right (x - h)
If "h"< 0 parabola shifts to the left ( x + h )

So, giiven y = (x-3)² +4 [which is basically, y= a(x-h) + k] the vertex would be (h,k) and because we know what "h" and "k" are 3, and 4, already we know the vertex is (3,4)

Domain = (-00,00) Range [4,00) <- The range is different for this equation. Axis of symmetry is x = h. Given y = -(x-1)² = 4
Vertex; (1,4)
AOS ; x =1
Domain ; (-00,00)
Range(-00, 4]
Direction of opening ; Downward

How to find Zeroes ;
First we factor out
y = -(x-1)(x-1)+4
Multiply the polynomials
= -(x²-2x+1) +4
Use the negative sign to change the sign
= -x²+2x-1+4
Put 0 in place of y because we're finding the zeroes.
0 = -x² +2x+3
Factor out the negative
0 = -(x²-2x-3)
Factor more
0 = - (x-3)(x-1)
Therefore x = 3, x =1, Existing zeroes are 3,1.

That's it.

Next scribe will be... duck, duck, Roxee.

posted by Rence @ 5:48 PM