x^2sk j =_, is=-*l -S~^3 SJ-^-K ~<==t M. Peacock Name Pecalculus Couse - Algeba II Review Packet This packet is to help you eview vaious topics that ae consideed to be peequisite knowledge upon enteing PeCalculus. Show all of you wok NEA TL Y and oganized! You may check solutions on you calculato, but be sue to show all wok fo cedit. Questions with NO wok will eceive NO cedit!! Box you answes! NO CALCULATOR UNLESS OTHERWISE STATED! I. Geomety Topics - Midpoint fomula: - Median of a A: A segment fom a vetex to the midpoint of the opposite side. - Angle Bisecto of a A: A segment fom a vetex which bisects the angle. - Pependicula Bisecto: A line passing though the midpoint of and pependicula to a segment. - Altitude of a A: A segment fom a vetex pependicula \ to the opposite side. Equations of a line: 1. Slope intecept: y=mx + b, v v whee slope= xj * x, - x, 2. Pointslope: y-y t =m(x-.t,) 3. Standad: - Distance fomula: Diections - State all linea equations in Standad Fom unless othewise stated. 1. Given AABC with A(-5,4), B(l, 6) and 2. Wite the equation of the line paallel to the line C(3, 8), wite the equation of the median 4x - 6y = -1 and contains the x-intecept of fom point C. 3. Wite the equation of the line, in slope intecept fom, though (2, -4) and pependicula to x - 2y = 1. 4.Find the value of "a" if a line containing the point 2 (a, -3a) has a y-intecept of 7 and a slope of --.
5. Given the distance between (x, 1) and 6. Wite the equation of the pependicula bisecto (-2, 5) is 2 7. Find the value(s) of x. of the segment joining A(-5, 4) and B(3, -6). Leave you answe in simplest fom. II. Quadatics A. Factoing - Stategies to ty when Factoing: - GCF - Guess and Check - Diffeence of two squaes - Gouping - Sum/Diffeence of cubes a 3 + b 3 = (a + b)(a 2 - ab + b 2 ) a 3 - b 3 = (a - b)(a 2 + ab + b 2 ) 1. Diections - Facto completely each of the following: a. 4x 2 +27x +35 b. -28y 2 + 7t 2 c. x 3-2x 2-9x +18 d. 8a 4 +27ab 3 B. Equations - Since the following ae equations, we can now go a step futhe and solve fo x by factoing o using the quadatic fomula. 2. Diections - Solve each of the following: a. -3x 2-5x +12 = 0 b. 3x 2 +5x = 6 c. x 2 +2x + 3 = 0
C. Gaphing - To gaph a quadatic equation in standad fom, y = ax 2 + bx + c, find the impotant points of the gaph by following the steps: Y-intecept: If a point is the y-intecept of the cuve, then that is the point at which the gaph cosses the y-axis. Since this point is on the y-axis, then the x- coodinate must be 0. Substitute zeo in fo x and solve fo y. Vetex: x-coodinate of the vetex: x = - b 2a. X-intecepts: y-coodinate of the vetex: substitute the value found fo the x-coo. into the oiginal equation and solve fo y. If a point is an x-intecept of the cuve, then it is a point at which the gaph cosses the x-axis. Since these points ae on the x-axis, then the y- coodinates must be 0. Substitute zeo in fo y and solve fo x by factoing o using the quadatic fomula. *No calculato, but you should also be able to gaph with the use of you calculato. 3. Diections - Given y = -3x 2-6x +2, find and gaph. a. y-intecept b. vetex c. x-intecepts III. Systems Substitution o Linea Combination (Elimination) can be used to solve systems of equations. - If thee is a solution to the system, then the equations ae epesenting intesecting lines. - If both vaiables cancel out and an equation is fomed that is neve tue, then thee is no solution and the lines neve intesect. Lines that neve intesect ae paallel lines. - If both vaiables cancel out and an equation is fomed that is always tue, then thee ae infinitely many solutions and the equations must epesent the same line. Diections - Solve each of the following. - Explain what the solution tells us about the lines epesented by the equations. - No calculato, but you need to be able to solve with the use of a calculato as well. 1. Ï 3x - 4 y = 2 Ì Ó - x + 3y = 1 2. Ï -x + y = 3 Ì Ó 2 x - 2y = -6 Solution: Explanation: Solution: Explanation:
IV. Exponents Diections - Simplify using only positive exponents and no calculato!!! Popeties: a m a n = a m +n a m a -n = 1 a n ( ) n p = a m n a Ê aˆ Ë b m = a m b m = a p a m a n = a m -n 1. Ê 81ˆ Ë 64-1 2 2. ( 27-2 ) - 1 3 3. ( 3x 2 ) -1 4. a. -2 4 6x -3 b. (-2) 4 5. 3-5 3 10 3 2 6. 4-1 + 2-1 ( ) 2 - hint 1: ( a -m + a -n ) p a -mp + a -np - hint 2: Apply the neg. exponent popety to each tem. Then get a common denom. and add V. Logaithms Given log b a = x, then b x = a whee b > 0 but b 1, and a > 0. Diections: - Solve fo x. 1. 3log 2 x = 12 2. log 5 125 = x 3. 3 + 4 log x 4 = 5 4. 3 2 log ( x + 5) = 1 27 5. 1 + 4 3 log ( x - 3) 4 = 11 3 6. log 5 25 4x-1 = 3
VI. Rational Expessions Diections - Simplify to a single faction: 1. Hint: get a common denominato! 2. Hint: facto and cancel! 1 ab - 2 b 2 x 2 + 6x + 8 x 2-4 3. Hint: get a common denominato in the numeato and multiply by the ecipocal x x - 1 +1 x + 2 x VII. Quick Gaphs: Diections - Gaph each of the following. - If you don t emembe, use you gaphing calculato to help you detemine the pattens. But you need to be able to do these gaphs without you calculato! 1. y = x - 2-3 2. y = ( x + 2) 2 + 1 3. y = x - 1 3 4. y = x +1-2 5. y = ( x - 3) 3 + 2 6. y = x + 3