Download PreCalc Unit 1 Day 3 Notes Big Ideas – Symmetry of graphs are

PreCalc Unit 1 Day 3 Notes
Big Ideas – Symmetry of graphs are suggested by the equations that create them.
Graphing relations of the form x = f (y) is analogous to graphing y = f (x).
Recall: Graphing functions quadratic in x.
A. 3 Forms:
1. Standard Form - f ( x)  ax 2  bx  c
 b
a. Vertex :   ,
 2a
 b 
f    
 2a  
b. y – intercept: (0, c)
c. vertical stretch - a
2. Vertex Form - f ( x)  ax  h   k
2
a. Vertex : h, k 
b. vertical stretch - a
3. Zeros Form - f ( x)  ax  z1 x  z 2 
a. x intercepts:
z1 ,0 and z2 ,0
 z  z 2  z1  z 2  
b. vertex:  1
, f
 
2
2



B. All forms create graphs that are symmetric with the line of symmetry x = h.
C. The symmetry and the range come from the fact that ultimately the y values
come from squaring numbers.
x
y  x2
3
9
2
4
1
1
0
0
1
1
2
4
3
9
Squaring 0 gives a unique
result. The square that results
from using x = 1 and x = -1
match, creating symmetry with
respect to x = 0.
Outputs never go below 0,
since any real number squared
is positive.
x
y   x  3  1
6
10
5
5
4
2
3
1
2
2
1
5
0
10
2
Squaring 0 gives a unique result.
(This occurs when we use x = -3).
The square that results from using
x = -4 and x = -2 match, creating
symmetry with respect to x = -3.
Outputs never go below 1, since
any real number squared is 0 or
more, and these results are having
1 added to them.
II. Equations that are quadratic in the variable y.
A. To graph x  f ( y ) , plug in y’s and get out x’s…
x   y  2   1
x  y2
1. Ex:
x
2
y
Vertex at (1, -2)
y values one above or
below that of the vertex
result in x’s that are 1 less,
since a = -1.
4 2
1
1
0
0
1
1
4
2
Examples: Find the vertex, x and y intercepts, line of symmetry and graph the following.
1)
x = y2 – 2y – 3
b
 b 
; h = f 

2a
 2a 
  2 
1
k=
2(1)
Vertex: k = 
 (-3, 2)
 (-4, 1)
 (-3, 0)
x = -3y2 + 6y + 24
2)
Vertex: k = 
k=
b
 b 
; h = f 

2a
 2a 
(24, 2) 
(27, 1) 
(24, 2) 
 (6)
1
2(3)
h = (1)2 – 2(1) – 3 = -4
h = -3(1) + 6(1) + 24 = 27
From (-4, 1), go up and down
1 and right 1.
From (27, 1), go up and down
1 and left 3.
x – int.: Plug y = 0. x = (0)2 – 2(0) – 3 . (-3, 0)
x – int. : Plug y = 0. x = -3(0)2 +6(0) +24 . (24, 0)
y-int: Plug x = 0. 0 = y 2  2 y  3
y-int: 0  3 y 2  6 y  24 (  -3)
0 = y2  2y  8
0 = (y – 4)(y + 2)
y = 4, y = -2
0 = (y – 3)(y + 1)
y = 3, y = -1
Line of Symmetry: y = 1 (passes through vertex)
Line of Symmetry: y = 1

x  11  (9)  1y
3) Rewrite x = -y2 – 6y – 11 in vertex form, then graph it.
x  11  1 y 2  6 y
2
 6y  9
x + 2 = -  y  3


2
x   y  3  2
2
 (-6, -1)
 (-3, -2)
 (-2, -3)
 (-3, -4)
 (-6, -5)
Vertex (-2, -3). Parabola opens to the left since a < 0.
(up and down 1, left 1 unit)
4) Graph x  2 y  52 y  1 . Give the coordinates of the vertex and 2 other points.
Each factor is associated with a y intercept. The vertex will have a
y value that is the midpoint of the y intercepts. We’ll plug that k
value into the function to find h.
 (0, 1/2)
 (-9, -1)
 (0, -5/2)
0 = 2 y  52 y  1
2y + 5 = 0
2y – 1 = 0
2y = -5
2y = 1
5
1
y
y
2
2
k

5 1

2 2   2  1
2
2
h = 2(1)  52(1)  1  (3)(3)  9
HW: PreCalc Unit 1 Day 3 HW