10/19/2006 Pre-Calculus polynomial function degree nlead coefficient 1 a zero function f(x) = 0 undefined constant function f(x) = 5 0 linear function f(x) = 2x + 5 quadratic function f(x) = x 2 + 2x slope linear constant non-zero

10/19/2006 Pre-Calculus roots or solutions x = -1 or 3.5 vertex: (h, k) complete the square vertex: (–4, –1) axis of symmetry: x = –4

10/19/2006 Pre-Calculus vertex: (1, 5) vertex: x – intercepts:

10/19/2006 Pre-Calculus constants power constant of variation or proportion varies as is proportional to power function power: –4 not a power function: power isn’t a constant constant of variation: 2 power function power is 1, constant of variation is 2  independent variable: r power is 2, constant of variation is 1 power: 2 direct variation constant of variation: 

10/19/2006 Pre-Calculus d = k F d = k t 2 monomial degree: 0 lead coefficient: 4 not monomial lead coefficient: 13 power is ½ (not an integer) not monomial monomial degree: 3 non-negative integer power is a variable

10/19/2006 Pre-Calculus vertical stretch / shrink reflection across the x-axis domain range continuity increasing decreasing symmetry boundedness extrema asymptotes end behavior

10/19/2006 Pre-Calculus dividend divisor quotient remainder

10/19/2006 Pre-Calculus k = 3 (3) 2 – 4(3) – 5 = 9 – 12 – 5 = –8 k = –2 (–2) 2 – 4(–2) – 5 = – 5 = 7 k = 5 (5) 2 – 4(5) – 5 = 25 – 20 – 5 = 0 divides evenly x - intercept zero solutionroot

10/19/2006 Pre-Calculus 3(x + 4)(x – 3)(x + 1) so factors are: x + 4, x – 3, x + 1 = 3x 3 + 6x 2 – 33x – 36 2(x + 3)(x + 2)(x – 5) so factors are: x + 3, x + 2, x – 5 = 2x 3 – 38x – 60

10/19/2006 Pre-Calculus (x + 4)(x – 4) = 0 f(x) = x 2 – 16 x = 4, x = –4 rational zeros

10/19/2006 Pre-Calculus potential: Use the rational zeros theorem to find the rational zeros of f(x) = 2x 3 + 3x 2 – 8x + 3 p = integer factors of the constant q = integer factors of the lead coefficient

10/19/2006 Pre-Calculus

10/19/2006 Pre-Calculus complex (real and non-real) zeros * non-real zeros are not x – intercepts zeros: 3i, – 3i, – 5 x-intercepts: – 5 complex conjugate (a + bi and a – bi)

10/19/2006 Pre-Calculus x 4 – 14x x 2 – 206x + 221

10/19/2006 Pre-Calculus denominator the x – axis ( y = 0 ) the line y = a n / b m there is no quotient output input

10/19/2006 Pre-Calculus vertical asymptote: horizontal asymptote: x – intercept y – intercept vertical asymptote: horizontal asymptote: x – intercept y – intercept none y = 0 none (0, 4) x = –1 none (0, 0) (1, 0) (0, 0) slant asymptote: y = x – 2

10/19/2006 Pre-Calculus (–3, 4) U (4,  ) because the graph crosses the x-axis because the graph does not cross the x-axis [ –3,  ) (– , –3)

10/19/2006 Pre-Calculus 1, –3, 2 (– , –3) U (1, 2) U (2,  ) (–3, 1) – –

10/19/2006 Pre-Calculus Write a standard form polynomial function of degree 4 whose zeros include 1 + 2i and 3 – i. quiz

10/19/2006 Pre-Calculus Solve the following inequality using a sign chart: x 3 + 2x 2 – 11x – 12 < 0 quiz

10/19/2006 Pre-Calculus Write the following polynomial function in standard form. Then identify the zeros and the x – intercepts. f(x) = (x – 3i) (x + 3i) (x + 4) zeros: x – intercepts: quiz

10/19/2006 Pre-Calculus Without graphing, using a sign chart, find the values of x that cause f(x) = (x – 2) (x + 6) (x + 1) to be: a.) zero ( f(x) = 0 ) b.) positive ( f(x) > 0 ) c.) negative ( f(x) < 0 ) a.) 2, –1, –6 b.) (–6, –1) U (2,  ) c.) (– , –6) U (-1, 2) quiz

10/19/2006 Pre-Calculus Use the quadratic equation to find the zeros of f(x) = 5x 2 – 2x + 5. Your answer must be in exact simplified form. quiz

10/19/2006 Pre-Calculus Find all zeros of f(x) = x 4 + 3x 3 – 5x 2 – 21x + 22 and write f(x) in its linear factorization form

10/19/2006 Pre-Calculus 2i is a zero of f(x) = 2x 4 – x 3 + 7x 2 – 4x – 4. Find all remaining zeros and write f(x) in its linear factorization form. quiz