Exponential Functions and Their Graphs, Logs and Natural Logs, Rational Functions and their Graphs

The exponential function f with base a is defined by f(x) = ax where a > 0, a  1, and x is any real number. For instance, f(x) = 3x and g(x) = 0.5x are exponential functions. Copyright © by Houghton Mifflin Company, Inc. All rights reserved

The Graph of f(x) = ax, a > 1 y Range: (0, ) (0, 1) x Horizontal Asymptote y = 0 Domain: (–, ) Copyright © by Houghton Mifflin Company, Inc. All rights reserved

The Graph of f(x) = ax, 0 < a <1 y Range: (0, ) Horizontal Asymptote y = 0 (0, 1) x Domain: (–, ) Copyright © by Houghton Mifflin Company, Inc. All rights reserved

Example: Sketch the graph of f(x) = 2x. x f(x) (x, f(x)) -2 ¼ (-2, ¼) y x f(x) (x, f(x)) -2 ¼ (-2, ¼) -1 ½ (-1, ½) 1 (0, 1) 2 (1, 2) 4 (2, 4) 4 2 x –2 2 Copyright © by Houghton Mifflin Company, Inc. All rights reserved

Example: Sketch the graph of g(x) = 2x – 1. State the domain and range. y f(x) = 2x The graph of this function is a vertical translation of the graph of f(x) = 2x down one unit . 4 2 Domain: (–, ) x y = –1 Range: (–1, ) Copyright © by Houghton Mifflin Company, Inc. All rights reserved

Example: Sketch the graph of g(x) = 2-x. State the domain and range. y f(x) = 2x The graph of this function is a reflection the graph of f(x) = 2x in the y-axis. 4 Domain: (–, ) x –2 2 Range: (0, ) Copyright © by Houghton Mifflin Company, Inc. All rights reserved

The irrational number e, where e  2.718281828… is used in applications involving growth and decay. Using techniques of calculus, it can be shown that Copyright © by Houghton Mifflin Company, Inc. All rights reserved

The Graph of f(x) = ex x f(x) -2 0.14 -1 0.38 1 2.72 2 7.39 y x 6 4 2 1 2.72 2 7.39 6 4 2 x –2 2 Copyright © by Houghton Mifflin Company, Inc. All rights reserved

Properties of Logarithmic Functions If b, M, and N are positive real numbers, b  1, and p and x are real numbers, then: Log15 1 = 0 150 = 1 Log10 10 = 1 101 = 10 Log5 5x = x 5x = 5x 3log x = x 3

Graph and find the domain of the following functions. y = ln x x y -2 -1 1 2 3 4 .5 cannot take the ln of a (-) number or 0 ln 2 = .693 ln 3 = 1.098 ln 4 = 1.386 D: x > 0 ln .5 = -.693

Graph y = 2x y = x x y -2 -1 1 2 2-2 = 2-1 = 1 2 4 The graph of y = log2 x is the inverse of y = 2x.

The domain of y = b +/- loga (bx + c), a > 1 consists of all x such that bx + c > 0, and the V.A. occurs when bx + c = 0. The x-intercept occurs when bx + c = 1. Ex. Find all of the above for y = log3 (x – 2). Sketch. D: x – 2 > 0 D: x > 2 V.A. @ x = 2 x-int. x – 2 = 1 x = 3 (3,0)

f (-x) = f (x): y-axis symmetry f (-x) = -f (x): origin symmetry 3.6: Rational Functions and Their Graphs Strategy for Graphing a Rational Function Suppose that where p(x) and q(x) are polynomial functions with no common factors. 1. Determine whether the graph of f has symmetry. f (-x) = f (x): y-axis symmetry f (-x) = -f (x): origin symmetry 2. Find the y-intercept (if there is one) by evaluating f (0). 3. Find the x-intercepts (if there are any) by solving the equation p(x) = 0. 4. Find any vertical asymptote(s) by solving the equation q (x) = 0. 5. Find the horizontal asymptote (if there is one) using the rule for determining the horizontal asymptote of a rational function. 6. Plot at least one point between and beyond each x-intercept and vertical asymptote. 7. Use the information obtained previously to graph the function between and beyond the vertical asymptotes.

EXAMPLE: Graphing a Rational Function 3.6: Rational Functions and Their Graphs EXAMPLE: Graphing a Rational Function Solution Step 1 Determine symmetry: f (-x) = = = f (x): Symmetric with respect to the y-axis. Step 2 Find the y-intercept: f (0) = = 0: y-intercept is 0. Step 3 Find the x-intercept: 3x2 = 0, so x = 0: x-intercept is 0. Step 4 Find the vertical asymptotes: Set q(x) = 0. x2 - 4 = 0 Set the denominator equal to zero. x2 = 4 x = 2 Vertical asymptotes: x = -2 and x = 2. more

EXAMPLE: Graphing a Rational Function 3.6: Rational Functions and Their Graphs EXAMPLE: Graphing a Rational Function Solution Step 5 Find the horizontal asymptote: y = 3/1 = 3. Step 6 Plot points between and beyond the x-intercept and the vertical asymptotes. With an x-intercept at 0 and vertical asymptotes at x = 2 and x = -2, we evaluate the function at -3, -1, 1, 3, and 4. -5 -4 -3 -2 -1 1 2 3 4 5 7 6 Vertical asymptote: x = 2 Vertical asymptote: x = -2 Horizontal asymptote: y = 3 x-intercept and y-intercept x -3 -1 1 3 4 f(x) = The figure shows these points, the y-intercept, the x-intercept, and the asymptotes. more