MATH 143 – Review Sheet for Test #1 – 9/21/05 1Last update: Tue Sep 20 08:09:47 MDT 2005 /m143.fa05/handouts143/t1 143 921/review suggestions 1 tex1 This list is not in final form. Like, stuff may yet be added to it.2 Test #1 isWednesday9/21/053 There will be a with-calculator part of this exam. So have some batteries in your calcu-lator. And bring it to the test.4 The test will cover the material of Assignments #1 – #15 roughly.5 Be sure you can(a) Add algebraic fractions using the Least Common Denominator(b) Derive the Quadratic Formula and be able to discuss the Discriminant.(c) Make a sign-change chart for a product of binomials.(d) That you know the two big triangle theorems:(i) Pythagoras’s Theorem(ii) The Similar-Triangles TheoremThese both relate picture information to algebra.These theorems lie at the base of all the straight-line-equations facts: slope worksbecause of the Similar-Triangles Theorem; the perpendicularity-and-slope criterionworks because of Pythagoras.The equation of a circle comes straight from the distance formula, which, in turn, is amanifestation of Pythagoras’s Theorem.(e) Completing the Square lies at the heart of(i) deriving the quadratic formula(ii) parsing a quadratic-in-x-and-y equation to see whether it’s a circle.(f) Add algebraic fractions using the Least Common Denominator(g) Decode negative exponents in expressions.(h) Decode fractional exponents in expressions.MATH 143 – Review Sheet for Test #1 – 9/21/05 2(i)(j)6 Re the absolute-value inequalities on assignment #14:(a) |x + 1| > 3.For this one, many folks offered this beginning ploy:−3 > x + 1 > 3,which excites suspicion because it implies that −3 > 3. So, don’t do that!Method AOne can correctly say that this inequality means that (x + 1) lies at least3 units from zero, so(i) (x + 1) > 3 or x > 2, OR(ii) (x + 1) 6 −3 or x 6 −4.We can put these two solutions together into the BOB answer:(−∞, −4] ∪ [2, +∞).Method BHere’s the trick presented in class: trade the given inequality in, temporarily,on the complementary inequality:|x + 1| < 3.This complementary inequality says that the distance from (x + 1) to 0 isless than 3, so−3 < (x + 1) < 3.We can add −1 to all three sides of this last inequality to get−4 < x < 2,which means that the complementary inequality’s solution is (−4, 2).This means that the original inequality, |x + 1| > 3, has solution set(−∞, −4] ∪ [2, ∞)(b) The second problem’s solution:−45,85.MATH 143 – Review Sheet for Test #1 – 9/21/05 3(c) The third problem’s solution: (−∞, −3] ∪ [−1, ∞).This arises from using algebra on the given inequality to change it to |2x + 4| > 2.7 The purple-page end-of-chapter-1 Test on page 132++ has all the BOB answers. All ofthe problems are fair game EXCEPT(i) 3, 9(ii) 19, 20, 21 (guaranteed to be on test