Mathematics

1. Chapter 1 Review
Functions and Their Graphs

2. The material in chapter 1 has
been covered in your previous
math courses. A summary is
listed on p.48. If you need more
tutorial review, try the links
at www.oldrochester.org/hs
Check the Precal PPTS.

3. y1
y2
x1 x2
(x1,y1)
(x2,y2)
lx2-x1l
ly2-y1l
a2
b2
d
c2

4. d  x2  x1
 
2
 y2  y1
 
2
a 2 + b2
c = √
Distance formula is based
on Pythagorean Theorem.

5. 1a Find distance between
(5, 1) and ( 2, -6 )
(x1, y1) (x2, y2)
d  x2  x1
 
2
 y2  y1
 
2
( 2 - 5)2 + ( -6 - 1 )2
d  x2  x1
 
2
 y2  y1
 
2

6. d  x2  x1
 
2
 y2  y1
 
2
( 2 - 5)2 + ( -6 - 1 )2
d = √
d = √
d = √
(- 3) 2 + ( - 7 )2
9 + 49
58 = 7.6
careful !
this is not
- 32
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QuickT ime™ and a
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QuickT ime™ and a
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QuickT ime™ and a
T IF F (Uncompressed) decom pressor
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QuickT ime™ and a
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7. QuickTim e™ and a TIFF (Uncompressed) decompressor are needed to see this picture.
(-3)2 = 9
-32 is NOT the same
Don’t forget
the ( ) !!!

8. 1b) For any 2 points (x1, y1)
and (x2, y2) the midpoint is
x1  x2
2
,
y1  y2
2





(x1,y1)
(x2,y2)

9. Find m (9, 4) (13, 12)
m = y2 - y1
x2 - x1
m = 12 - 4
13 - 9
= 8
4
= 2
1c

10. 1d) Find an equation for a
vertical line through (2, 4 ).
2) Vertical line
has undefined
slope.
X = 2

11. Why are the lines parallel ?
Same slope.
m = 2
y1 = 2 x + 1
y2 = 2 x - 3
3)

12. y = 3/4 X + 2
y = - 4/3 X
slopes have
opposite signs
& are reciprocals
 lines
On your
calculator
use ( ) on
fractions
Perpendicular
Lines
3

13. Forms of Linear Equations
General or Standard
Ax + By = C
Slope Intercept
y = mx + b
4)

14. Forms of Linear Equations
Point Slope Form
y - y1
x - x1
=m or y -y1= m(x - x1)
Intercept Form
x
a
+
y
b
=1

15. Function - for each X
there is only one Y.
Zero of the function is
where the graph
crosses the
X axis.
5)

16. Imaginary Numbers
Define
i 2 = -1 and √-1 = i
√- 9
no answer in Reals
6a)

17. 6b Form of a complex number
a + bi (a and b are reals. i is imaginary
The conjugate is a - bi .
( a + bi ) (a - bi ) =
a2 - (bi )2 =
a2 - b2 (-1) =
a2 + b2
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18. Reals and imaginary numbers
are subsets of
COMPLEX NUMBERS.
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19. Solving Quadratics
Ax2 + Bx + C = Y
7

20. Quadratics
Solve by factoring.
x2 - 4 = 0
(x+2)(x-2) = 0
x+2 = 0 or x - 2 = 0
x = -2 x = 2

21. Solve with Quadratic Formula
4 x2 – 8 x + 1 = 0
a b c
a
ac
b
b
x
2
4
2




      
 
4
2
1
4
4
8
8
2






x
Careful when
substituting negatives

22. 8
48
8 

x
x= 8 + 6.9
8
x = 8 + 6.9
8
x = 8 - 6.9
8
or
=1.86 =.14

23. Solve by completing the square.
x2 + 6x -7 = 0
1) Be sure the coefficient of
the x2 term is 1.
2) Get the constant on the other side.
+7 +7
x2 + 6x = 7

24. Solve by completing the square.
x2 + 6x -7 = 0
3)Take half the coefficient of
the x term .Square it.
Add to both sides.
x2 + 6x = 7
6 / 2 =3
32 = 9
x2 + 6x +9 = 7+9

25. Solve by completing the square.
x2 + 6x -7 = 0
4) Factor the left.
Simplify the right.
x2 + 6x = 7
x2 + 6x +9 = 7+9
(x + 3)2 = 16
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26. Solve by completing the square.
x2 + 6x -7 = 0
5) Take the square root both sides.
x2 + 6x = 7
x2 + 6x +9 = 7+9
(x + 3)2 = 16
√ +√
(x + 3 ) = +4
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27. Solve by completing the square.
x2 + 6x -7 = 0
6)Solve
x2 + 6x = 7
x2 + 6x +9 = 7+9
(x + 3)2 = 16
√ +√
(x + 3 ) = +4
x+3 = 4 or x +3 = -4
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28. Solve by completing the square.
x2 + 6x -7 = 0
6)Solve
x2 + 6x = 7
x2 + 6x +9 = 7+9
(x + 3)2 = 16
√ +√
(x + 3 ) = +4
x = 1 or x = -7
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29. Quadratics
Solve with a graph.
X2- 4 = 0
graph
X2- 4 = Y
(-2,0) (2,0)

30. Y = Ax2 + Bx + C
Y = 2x2 - 4x - 1
Vertex (-b/2a, )
-b
2a
-(-4)
2(2)
= = 1

31. Y = Ax2 + Bx + C
Y = 2x2 - 4x - 1
Vertex (-b/2a, )
( 1 , )
substitute
(1, -3)

32. Y = Ax2 + Bx + C
Y = 2x2 - 4x - 1
vertex (1, -3)
line of symmetry
X = 1

33. Vertex Form
y = a(x -h)2 + k
y = 2(x -1)2 - 2
vertex (h, k )

34. thought for the day
Mistakes are
the portals
of discovery.