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Exam • May 15th 6:15pm. Be there early • Exam rooms: On website – http://www.math.ksu.edu/math100/ • Exam rooms are by recitation instructor (not me) • Bring your k-state student ID How to study • Focus on your previous exams. – Review the problems and your own work • Study Guide – http://www.math.ksu.edu/math100/spring2009/FinalExamStudyGuideSpring09.pdf • Take old finals (in realistic conditions) – http://www.math.ksu.edu/course_info/oldtests/1 00tests/ Your test • 75% questions from your previous tests this semester with the numbers changed • 25% new material – Inequalities – Composition – Exponents & logs – Systems • STUDY YOUR OLD TESTS Agenda • Today – Functions – Linear problems (equalities, inequalities, systems) – Polynomials (including quadratics) • Thursday – Radicals – Rationals – Exponents and logs Functions A function is a relationship between two changing variables • An “input” variable • An “output” variable – The result of “doing” the function to the output variable • Both variables change so that the “input” variable always tells you exactly what the “output” variable is. – You never get two outputs for the same input. Not a Function output input Intercepts y x-intercept (-2,0) x-intercept (2,0) x y-intercept (0,-4) Combining Functions • • • • • (f+g)(x)=f(x)+g(x) (f-g)(x)=f(x)-g(x) (fg)(x)=f(x)*g(x) (f/g)(x)=f(x)/g(x), g(x)≠0 (f∘g)(x)=f(g(x)) In picture form f(x) f x * g f(x)g(x) g(x) Is not the same as g x f g(x) f(g(x)) COMPARISON f (x) = 2x -1, g(x) = (x - 3)2 (f∘g)(3)=f(g(3)) (fg)(3)=f(3)g(3) g(3) = (3- 3)2 g(3) = 0 ƒ(g(3)) = ƒ(0) = 2 * 0 -1 = -1 ƒ(g(3)) = -1 -1≠0 f (3) = 5 g(3) = 0 f (3)g(3) = ( 5) ( 0) = 0 ( fg)(3) = 0 Graphing Transformations • The graph for ƒ(x)+c is the graph of ƒ(x) shifted up by c. • The graph for ƒ(x-a) is the graph of ƒ(x) shifted right by a. – NOTE THE MINUS SIGN • The graph for rƒ(x) is the graph of ƒ(x) stretched vertically by r. – negative r causes the graph to flip vertically. • The graph for ƒ(sx) is the graph of ƒ(x) squished horizontally by s. – Negative s causes the graph to flip horizontally • Note the difference! Even and Odd • A function ƒ is EVEN if ƒ(-x)=ƒ(x). Example: x2 • A function ƒ is ODD if ƒ(-x)=-ƒ(x). Example: x3 • A function ƒ is NEITHER if ƒ(-x)=something else. Example: x3+1 Function inverse Cubing to cube root y=x3 x=∛y Cubing to cube root The relationship between x and y stays the same Only my point of view changes y=x3 x=∛y How to find a function inverse • • • • • ƒ(x)=………….x…………. Rewrite as y=……………x………… Solve for x. x=~~~~y~~~~~~ Rewrite as an inverse ƒ-1(y)=~~~~y~~~~~~ OPTIONAL: change ys to xs. • ƒ-1(x)=~~~~x~~~~~~ • WARNING: Always check that your inverse is actually a function. Given f (x) = 7x + 1 on the domain of all real numbers, find f -1(x). Be sure to write your answer as a function of x. a) b) c) d) e) f -1(x) = (1/7)x − 1/7 f -1(x) = x − 1/7 f -1(x) = 1/(7x+1) Both (a) and (c) None of the above Given f (x) = 7x + 1 on the domain of all real numbers, find f -1(x). Be sure to write your answer as a function of x. ƒ(x)=7x+1 y=7x+1 (y-1)/7=x x=(1/7)y-(1/7) ƒ-1(y)=(1/7)y-(1/7) ƒ-1(x)=(1/7)x-(1/7) A Lines Point slope form • The equation of a line with slope m through point (a,b) is y - b = m(x - a) • If you don’t know the slope, know two points (a1,b1) and (a2,b2), then the slope m is just the slope formula for those points. b2 - b1 m= a2 - a1 Slope intercept form y = mx + b • Slope intercept form is the “simplest” form of a line – “Simplify” means put in slope intercept form Doing the same thing to both sides • Adding, Subtracting, Multiplying, Dividing, a number from both sides of the equation. – Changes the value of both sides, but not the equality. 12 = 12 - 6 = -6 6=6 4x = 4 + 2x - 2 x = -2 x 2x = 4 /2=/2 /2=/2 3=3 x=2 WARNING • Each side is a number. When multiplying (or dividing) multiply (or divide) the whole number. GOOD BAD GOOD 2+4x = 6 2+4 = 6 2+4 = 6 2 + 4 / 2 = 6 / 2 (2 + 4) / 2 = 6 / 2 (2 + 4x) / 2 = 6 / 2 2 / 2 + 4x / 2 = 3 2+2 =3 6/2=3 1+ 2x = 3 4=3 3=3 x =1 Rearranging an equation to solve 4 ( x 1) 8 2 x 4x 4 8 2x 4 4 4x 4 2x 2 x 2 x 2x 4 x2 Solving Inequalitites • When I divide by a negative, I can have the same effect as “moving to the other side” by “flipping the sign” Moved x 8 < 4 - 2x 8 + 2x < 4 8 < 4 - 2x 8 - 4 < -2x 2x < -4 4 < -2x x < -2 -2 > x Answers Match Flipped the sign Solving a system of equations on your calculator (and showing work) • Solve 4x + 8y - 4z = 8 2x + 3y + 4z = 4 5x + 8y + 1z = 7 In my calculator, I set the matrix [A] é 4 8 -4 8 ù ê ú [ A] = ê 2 3 4 4 ú ê 5 8 1 7 ú ë û Then I used the command rref([A]) The calculator output was So the answer is x=-3.5 y=3 z=0.5 é 1 0 0 -3.5 ù ê ú ê 0 1 0 3 ú ê 0 0 1 0.5 ú ë û Polynomials Arithmetic on complex numbers • 1 and i cannot be combined. They are on separate axes. – 1+i can’t be simplified, just like x+y can’t be simplified. • Treat i like a variable and you will be ok. • Remember that i2=-1 and √(-1)=i – This can be simplified 2 + 3i 1+ 2i Examples (2 + 3i)(1- 2i) = 1- (-4) (2 + 3i) (1- 2i) = (1+ 2i) (1- 2i) (2 + 3i)(1- 2i) = 5 (2 + 3i)(1- 2i) = 12 - (2i)2 8-i = 5 You are not done until you have the real and imaginary parts completely separate 8 1 = - i 5 5 Two Formulas Quadratic Formula -b b 2 - 4ac x= ± 2a 2a 2 When ax + bx + c = 0 Vertex Formula -b h = , k = f (h) 2a 2 When f (x)=ax + bx + c Vertex is (h, k) Vertex form • y=a(x-h)2+k • To find an equation of a parabola from vertex (h,k) and point (x1,y1). – Plug in h,k, x1,y1 and solve for a. – Plug in h,k, and a. – Answer should look something like: y=2(x-1)2-2 Standard Form of a Polynomial 3x2+2x-2x4-3 Constant term = y-intercept Leading Term 3x2+2x-2x4-3 Leading Coefficient determines end behavior Degree = number of roots = number of bends +1 Factored Form of a Polynomial 4(x-4)(x--1/3)(x-i)(x--i) Leading Coefficient End Behavior (+ means y is increasing for big x. – means y is decreasing when x is big) Roots Solving Polynomials 1) Use the Rational Root Test to come up with guesses for roots. 2) Use Synthetic Division to test roots and factor the polynomial 3) When you have only a quadratic left, use the quadratic formula Rational Root Test 4x3-3x2+2x-5 Factors of -5: {-5,-1,1,5} Factors of 4: {1,2,4} The only possible rational roots of this polynomial are -5/1, -1/1, 1/1, 5/1, -5/2, -1/2, 1/2, 5/2, -5/4, 1/4, 1/4, 5/4 Review: Synthetic Division • x3+x2-4x-4. Root at x=2 1 3 2 0 2 | 1 1 -4 -4 2 6 4 Add up Multiply to the bottom Add up Multiply to the bottom Add up Multiply to the bottom Add up 1x3+1x2-4x-4=(x-2)(1x2+3x+2)+0 You are given the coordinates of the vertex (-8,3) and of a point (-4,7) on a parabola. Find the equation of the parabola. a) y = -.25(x+8)2 - 3 b) y = .25(x+8)2 - 3 c) y = .25(x-8)2 + 3 d) y = .25x2 + 4x + 19 e) Both (c) and (d) You are given the coordinates of the vertex (-8,3) and of a point (-4,7) on a parabola. Find the equation of the parabola. 7=a(-4- -8)2+3 7=a(4)2+3 4/42=a a=1/4 y=0.25(x+8)2+3 y=0.25(x+8)(x+8)+3 y=0.25x2+4x+19 D Solving Polynomial Inequalities • • • • x2<x3-3x Get 0 on one side [x2-x3+3x<0] Graph the polynomial [y=x2-x3+3x] Convert to an equality [x2-x3+3x=0] Find the roots x=0, x=0.5+0.5√(13), 0.5-0.5√(13) • Use the roots and the graph to solve the inequality 0.5+0.5√(13)<x<0 OR x>0.5-0.5√(13) Test each interval Is (3)2+2(3)-(3)3<0 ? Is -12<0 ? YES. <----------------|------------|--------------------|---|----> -1 0 2 3 x<-1 -1<x<0 0<x<2 x>2 NO YES NO YES Solve the QUADRATIC INEQUALITY: Hint: You might graph the parabola y=(x-3)(x+4) first a) x > - 3 b) x < 4 c) x > 3 or x < - 4 d) - 4 < x < 3 e) None of the above C) x > 3 or x < - 4