Concepts: definition of polynomial functions, linear functions tree representations), transformation of y = x to get y = mx + b, quadratic functions axis of symmetry, vertex, x-intercepts), transformations of y = x 2 to get vertex form y = ax ) 2 + k, sketces Polynomial Functions Let n be a nonnegative integer and let a 0, a 1,,a n 1, a n be real numbers wit a n 0 Te function given by fx) = a n x n + a n 1 x n 1 + + a 1 x + a 0 is a polynomial function of degree n Te leading coefficient is a n Linear Functions Linear functions are discussed in Sections P3 and P4 in te prerequisites Tree different ways of writing te equation of a line: slope-intercept form, were m and y-intercept 0, b) are given: y = mx + b slope-point form, were m and point on te line x 1, y 1 ) are given: y y 1 = mx x 1 ) CALCULUS) point-point form, were two points on te line x 1, y 1 ) and x 2, y 2 ) are given: y y 1 = y 2 y 1 x x 1 ), wic is sometimes written as y y 1 = x x 1 x 2 x 1 y 2 y 1 x 2 x 1 Coose one of te above to work wit depending on wat information you are given formulas, altoug you could always work from y = mx + b if you like I suggest memorizing all tree Basic Function: Identity Function fx) = x Tis is a linear function wit slope m = 1 and y-intercept 0 Domain: x R Range: y R Continuity: continuous for all x Increasing-decreasing beaviour: increasing for all x Symmetry: odd Boundedness: not bounded Local Extrema: none Horizontal Asymptotes: none Vertical Asymptotes: none End beaviour: lim x = and x lim x = x Page 1 of 5
Transformations of Identity Function Basic function: y = x Vertical stretc of m units m > 1): y = mx Sift up b units b > 0): y = mx + b Example Write an equation for te linear function f suc tat f 1) = 2 and f3) = 2 Te two points te function must pass troug are x, fx)) = 1, 2) = x 0, y 0 ), and x, fx)) = 3, 2) = x 1, y 1 ) Te linear function is given by y = fx) = ax + b, or in point slope form by y y 1 = mx x 1 ) Te slope of te line is m = a = rise/run = 2 2))/ 1 3) = 4/ 4) = 1 Use te point-slope form of te line y 2) = 1)x 3) y = x + 3 2 = x + 1 Te linear function is given by fx) = x + 1 Alternate solution: Te two points te function must pass troug are x, fx)) = 1, 2) = x 0, y 0 ), and x, fx)) = 3, 2) = x 1, y 1 ) Te linear function is given by y = fx) = ax + b, or in point-point form by y y 1 = y 0 y 1 x x 1 x 0 x 1 y y 1 = y 0 y 1 x x 1 x 0 x 1 y 2) = 2 2) x 3 1 3 y + 2 x 3 = 1 y + 2 = x 3) y = x + 3 2 = x + 1 Te linear function is given by fx) = x + 1 Page 2 of 5
Quadratic Functions A quadratic function is a polynomial of degree 2, y = ax 2 + bx + c It takes tree points to determine te tree constants a, b, and c in a quadratic function Quadratic functions are described by teir axis of symmetry and vertex Basic Function: Squaring Function fx) = x 2 Domain: x R Range: y [0, ) Continuity: continuous for all x Increasing-decreasing beaviour: decreasing for x < 0, increasing for x > 0 Symmetry: even axis of symmetry is te line x = 0) Boundedness: bounded below Local Extrema: minimum at 0, 0) te vertex) Horizontal Asymptotes: none Vertical Asymptotes: none End beaviour: lim x x2 = and lim x x2 = Transformations of Squaring Function Basic function: y = x 2 Sift rigt by units > 0): y = x ) 2 Stretc vertically by a units a > 1): y = ax ) 2 Sift up k units k > 0): y = ax ) 2 + k Te Vertex Form Te form fx) = ax ) 2 + k is called te vertex form for a quadratic function Te vertex of te parabola is, k) Te axis of symmetry is x = If a > 0, te parabola opens up, if a < 0 te parabola opens down Page 3 of 5
x-intercepts Te x-intercepts of te quadratic function are found from te solution to te quadratic equation: fx) = ax 2 + bx + c = 0, wic can be determined using completing te square cf page 46) as x-intercepts = b ± b 2 4ac 2a Example Find te axis of symmetry and vertex of te quadratic function fx) = 3x 2 + 5x 4 by writing in te vertex form Ten find te x-intercepts and sketc te function To solve tis problem we need to write te quadratic function in vertex form We can do tis by completing te square 3x 2 + 5x = 3 x 2 + 53 ) x = 3 x 2 + 5 ) ) 2 ) 2 5 5 3 x + 3 6 6 fx) = 3x 2 + 5x 4 4 144 36 219 = 3 x 5 6 )) 2 73 12 From tis form we can identify te vertex as 5/6, 73/12), and te axis of symmetry as x = 5/6 Te quadratic function opens up since te leading coefficient is 3 > 0 Te x-intercepts are found using te quadratic formula to solve 3x 2 + 5x 4 = 0, or since we ave te vertex form we can get tem directly: 3 x + 5 ) 2 73 = 0 6 12 x + 5 ) 2 = 73 x + 5 73 = ± x = 5 6 ± 73 6 = 5 ± 73 6 Page 4 of 5
Te Average Rate of Cange Te average rate of cange of te linear function fx) = mx + b over te interval x, x + ) is te slope m average rate of cange = fx + ) fx) = mx + ) + b) mx + b) Te average rate of cange ere does not depend on te interval x, x + ) = mx + m + b mx b Te average rate of cange of te quadratic function fx) = ax 2 + bx + c on te interval x, x + ) is: fx + ) fx) average rate of cange = = ax + )2 + bx + ) + c) ax 2 + bx + c) = ax 2 + a 2 + 2ax + bx + b + c ax 2 bx c = a2 + 2ax + b a = + 2ax + b) = a + 2ax + b Notice tat te average rate of cange ere depends on te interval x, x + ) = m = m Page 5 of 5