Chapter 1 Review Functions and Their Graphs

tutorial review, try the links 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.

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

Distance formula is based on Pythagorean Theorem. √ c = a 2 + b2

(5, 1) and ( 2, -6 ) (x1, y1) (x2, y2) ( 2 - 5)2 + ( -6 - 1 )2 1a Find distance between (5, 1) and ( 2, -6 ) (x1, y1) (x2, y2) ( 2 - 5)2 + ( -6 - 1 )2

careful ! this is not - 32 ( 2 - 5)2 + ( -6 - 1 )2 d = √ (- 3) 2 + ( - 7 )2 9 + 49 58 careful ! this is not - 32 d = √ d = √ = 7.6

(-3)2 = 9 -32 is NOT the same Don’t forget the ( ) !!!

and (x2, y2) the midpoint is 1b) For any 2 points (x1, y1) and (x2, y2) the midpoint is (x1,y1) (x2,y2)

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

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

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

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

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

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

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

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

a + bi (a and b are reals. i is imaginary) 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

Reals and imaginary numbers are subsets of COMPLEX NUMBERS.

7 Solving Quadratics Ax2 + Bx + C = Y

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

substituting negatives Solve with Quadratic Formula 4 x2 – 8 x + 1 = 0 a b c Careful when substituting negatives

x= 8 + 6.9 8 x = 8 + 6.9 8 x = 8 - 6.9 8 or =1.86 =.14

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

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

x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 Solve by completing the square. x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 4) Factor the left. Simplify the right.

x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 Solve by completing the square. x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 √ +√ (x + 3 ) = +4 5) Take the square root both sides.

x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 Solve by completing the square. x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 √ +√ (x + 3 ) = +4 x+3 = 4 or x +3 = -4 6)Solve

x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 Solve by completing the square. x2 + 6x -7 = 0 x2 + 6x = 7 x2 + 6x +9 = 7+9 (x + 3)2 = 16 √ +√ (x + 3 ) = +4 x = 1 or x = -7 6)Solve

Quadratics Solve with a graph. X2- 4 = 0 (-2,0) (2,0) graph X2- 4 = Y

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

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

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

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

thought for the day Mistakes are the portals of discovery.