Algebra Review Exponents bx exponent Base Add exponent

Algebra Review: Exponents: bx exponent Base Add exponent terms: Just Combine Like Terms 2 x 3 + 5 x 3 = 7 x 3 Product Rule Multiply Like Bases, add exponents. Division Rule Divide Like Bases, subtract exponents. Power Rule 1 Negative exponents Fractional exponents Zero exponent Rule

Factoring: Steps: 0 Write the terms in descending order. 1 Always check for a GCF( Leftovers ) 2 TWO terms leftover. A. Difference of Perfect Squares x 2 – y 2 = ( x + y )( x – y ) B. Perfect Cubes 3 THREE terms leftover. ax 2 + bx + c a, b, c method. Handout in CANVAS 4 FOUR terms leftover. Factor by Grouping ___ + ___ __( __ + __ ) + __( __ + __ ) red ( )’s must be the same ( __ + __ )

3 4 ( 2 x + 5)( 2 x – 5 ) ax 2 + bx + c ___ * ___ = ac = 2(-12) 3 =b =5 8 – ___ even Answer Looks Like: Remove the GCF : odd ( 2 x + ___)( 2 x – ___) 2 The Answer is… ( x + 4 )( 2 x – 3 ) odd

Completing the Square/Quadratic Formula: 1

a b c

Solving Equations:

x

Geometry Review: Distance and Midpoint: EX #1: Find the distance and midpoint between (1, -2) and (5, 3).

Lines: Standard Form Ax + By = C m = -A/B

EX #2: a. ) Find the equation of the line through (1, -7) with a slope of -1/2.

b. ) Find the equation of the line through the points (-1, 2) and (3, -4).

EX #3: Sketch the graph of the equation: 3 x – 5 y = 15

Parallel and Perpendicular Lines: EX #4: Find the equation of the line through the point (5, 2) that is parallel to the line 4 x + 6 y + 5 = 0. EX #5: Show that the lines 2 x + 3 y = 1 and 6 x – 4 y – 1 = 0 are perpendicular

Circles: EX #6: a. ) Find an equation for the circle that has center (-1, 4) and passes through the point (3, -2). (h, k) (x, y)

TRUE f (-2) = 1 – (-2)2 = 1 – 4 = -3 1 > 0 TRUE f (1) = 2(1) + 1 = 3

Domain: The set of all x coordinates that generate y coordinates. Domain Test x values that are positive, negative, and zero with all the operators in the function. Adding, subtracting, multiplying and squaring work with +, -, and 0. However, dividing by 0 doesn’t exist, so set the denominator not equal to zero and solve for x to find the restriction. We can take the cube root of +, -, and 0. Another fraction with x in the bottom means find the restrictions. x 2 must be positive, so all x values will work. We can only take the square root of + values, and 0. Set the radicand > 0 and solve for x.

Intercepts: Order pair points where the graph crosses the x and y axis. To find x-intercept ( ___, 0 ) substitute 0 for y and solve for x. To find y-intercept ( 0, ___ ) substitute 0 for x and solve for y. x-intercept = ( ___, 0 ) This trinomial factors. Solving techniques are factoring, complete the square, or quadratic formula. y-intercept = ( 0, ___ ) is the y-intercept. are the x-intercepts.

Graphs: Two techniques. Find the intercepts. Convert to y = mx + b. x-int. (__, 0) & y-int. (0, __) 3 x = 12 & - 6 y = 12 x=4 y=-2 ( 4, 0 ) ( 0, - 2 ) - 3 x = - 3 x - 6 y = - 3 x + 12 -6 -6 -6 y=½x– 2 No y variable in the equation, so the line can not cross the y axis.

Translations: Standard notation for all functions y = a f (bx – c) + d. Order of transformations will be our rules for Order of Operations. 1. b : If b is negative flip over y-axis. | b | > 1, horizontal shrink | b | < 1, horizontal stretch 3. a : If a is negative flip over x-axis. | a | > 1, vertical stretch | a | < 1, vertical shrink 2. b x – c = 0 : Set equal to zero and solve for x. If x = a positive x, shift to the right x units If x = a negative x, shift to the left x units 4. d : Take for face value If d is positive, shift up d units If d is negative, shift down d units

Standard quadratic notation or vertex form of the quadratic. ( h, k ) is the vertex and a describes the pattern of vertical change. Pattern of vertical change starts at the vertex and is as follows: 1 right up/down 1 a 1 right up/down 3 a 1 right up/down 5 a. . .

2&3 1 4 Transformation Order. 1. x = 4, shift right 4 units. 2. a = -2, negative flips over x - axis. 3. | - 2 | = 2 > 0, vertical stretch by a factor of 2. 4. k = 5, shift up 5 units. GRAPH by hand. Vertex is at (4, 5) and the pattern 1 a, 3 a, 5 a, …with a = -2 wil be 1 rt down 2, 1 rt down 6, 1 rt down 10

2 1 3 Transformation Order. 1. x = - 3, shift left 3 units. 2. a = , vertical shrink by a factor of. Multiply the red y-coord. by a third. 3. d = - 2, shift down 2 units.

Consider the function Graph. on the graph. 1. Negative on the 2 will flip the graph over the y-axis and | -2 | = 2 will cause a horizontal shrink by dividing the x-coordinates by 2. 1. A quicker way is to divide all x -coordinates by -2. (-3, 2) (-1, 1) Graph . (-1. 5, -1) (1. 5, 2) (0. 5, 1) (1, -1) (-0. 5, -1) (3, -1)

Consider the function Graph. on the graph. (-6, 4) (-6, 2) Graph . 1. The “+3” on the inside of the ( )’s will move every x-coordinate to the left 3 units. 2. The multiplication of on the outside will multiply 2 to every y-coordinate. (-4, 2) (-4, 1) (-2, -1) (-3, 2) (-1, 1) ( 0, -1) (-2, -2) ( 0, -2) (1, -1) (3, -1)

Polynomial form Horizontal Asymptote Simplified Factored form Determined by the degree of the polynomial form, m & n Case 1: If m < n, then H. A. is at y = 0. Case 2: If m = n, then H. A. is at y = a/b Case 3: If m > n, then no H. A. If m is one degree bigger, then there is a slant asymptote. y-intercept Determined in Polynomial form, ( 0, c/d ). Vertical Asymptote Determined by Simplified Factored form, set Q(x) factors = 0 and solve for x. x-intercept Determined by Simplified Factored form, set P(x) factors = 0 and solve for x. Holes Determined by Simplified Factored form, the binomial factors that canceled is the x coordinate of the Hole.

P. F. S. F. F. 1 Hole @ x = 2 ( 2, 1 ) H. A. Case 1: y = 0 y - int. ( 0, -12/-8 ) = ( 0, 3/2 ) = ( 0, 1. 5 ) V. A. x+4=0 x=-4 x - int. None, no x’s left in numerator. Graphing Order x concepts 1 st. V. A. x = - 4 x - int. , None y concepts 2 nd H. A. y = 0 y - int. ( 0, 1. 5 ) Hole 3 rd ( 2, 1 )

1 H. A. Case 2: y = 2 y - int. ( 0, -8/-2 ) = ( 0, 4 ) V. A. x - int. Graphing Order x concepts 1 st. V. A. x - int. y concepts 2 nd H. A. y = 2 y - int. ( 0, 4 )

Trigonometry Review: Angles

Length of an Arc and Area of a Sector: EX #2: a. ) If the radius of a circle is 5 cm, what angle is subtended by an arc of 6 cm? b. ) If a circle has a r = 3 cm, what is the length of an arc & sector area subtended by a central angle of 3π/8 rad?