Copyright © 1999 by the McGraw-Hill Companies, Inc. Barnett/Ziegler/Byleen College Algebra, 6 th Edition Chapter Three Graphs & Functions

(a)Symmetry with respect toy axis (b)Symmetry with respect tox axis Symmetry

Symmetry (c)Symmetry with respect to origin (d)Symmetry with respect toy axis, x axis, and origin

Distance Between Two Points

Standard Equation of a Circle 1.Circle with radius r and center at ( h, k ): ( x – h ) 2 + ( y – k ) 2 = r 2 r > 0 2.Circle with radius r and center at (0, 0): x 2 + y 2 = r 2 r > 0 Circle

Line Slope Example Geometric Interpretation of Slope

Standard form Ax + By = CA and B not both 0 Slope-intercept form y = mx + b Slope: m ; y intercept: b Point-slope form y – y 1 = m ( x – x 1 )Slope: m ; Point: ( x 1, y 1 ) Horizontal line y = b Slope: 0 Vertical line x = a Slope: Undefined Equations of a Line

Vertical Line Test for a Function (a) y 3 – x = 1 (b) y 2 – x 2 =

x f(x) 5–5 10 –10 f(x) = –x 3 0 x g(x) 5–5 5 g(x) = 2x (a) Decreasing on (–   ) (b) Increasing on (– ,  ) Increasing, Decreasing, and Constant Functions

x h(x) 5–5 5 h(x) = 2 0 x p(x) 5–5 5 p(x) =x 2 – 1 Increasing, Decreasing, and Constant Functions (c) Constant on (–   ) (d)Decreasing on (– , 0] Increasing on [0,  )

Properties of a Quadratic Function and Its Graph f ( x ) = ax 2 + bx + c = a ( x - h ) 2 + k a  0 1.The graph off is a parabola:

5–5 5 x Identity Function f(x) f(x) = x 5–5 5 x Absolute Value Function g(x) = |x| g(x) Six Basic Functions

Six Basic Functions 5–5 5 Square Function x h(x) = x 2 h(x) 5–5 5 Cube Function x m(x) = x 3 m(x)

Six Basic Functions 5 5 Square-Root Function x n(x) =x n(x) 5–5 5 Cube-Root Function x p(x) = 3 x p(x)

Given functions f and g, then f ° g is called their composite and is defined by the equation (f ° g)(x) = f [g (x)] The domain of f ° g is the set of all real numbers x in the domain of g where g(x) is in the domain of f. Composite Functions

x y 5–50 5 (1, 1) x y 5–50 5 (1, 3) x y 5–50 5 (1, –2) (b) F: y = x 2 + 2(c) G: y = x 2 – 3 The graph of y = x is the same The graph of y = x 2 – 3 is the same as the graph of y = x 2 shifted up as the graph of y = x 2 shifted down two units. three units. (a) f: y = x 2 Vertical Shifts

x y 5–50 5 (1, 1) x y 5–50 5 (–1, 1) x y 5–50 5 (4, 1) (a) f: y = x 2 The graph of y = (x + 2) 2 is the sameThe graph of y = (x – 3) 2 is the same as the graph of y = x 2 shifted to theas the graph of y = x 2 shifted to the left two units.right three units. (b) P: y = (x + 2) 2 (c) Q: y = (x – 3) 2 Horizontal Shifts

(a) f: y = x 2 (b) R: y = –x 2 Reflection (c) S: y = 2x 2 Expansion (d) T: y =  x 2 Contraction Reflection, Expansion, and Contraction

(a) f(a) = f(b) for a  b; (b)Only one point has ordinate f(a); f is not one-to-onef is one-to-one A function is one-to-one if no two ordered pairs in the function have the same second component and different first components. Horizontal Line Test A function is one-to-one if and only if each horizontal line intersects the graph of the function in at most one point. One-to-One Functions

If a function f is increasing throughout its domain or decreasing throughout its domain, then f is a one-to-one function. (a) An increasing function is always one-to-one (b) A decreasing function is always one-to-one (c) A one-to-one function is not always increasing or decreasing Increasing and Decreasing Functions

Step 1.Find the domain of f and verify that f is one-to-one. If f is not one- to-one, then stop, since f –1 does not exist. Step 2.Solve the equation y = f(x) for x. The result is an equation of the form x = f –1 (y). Step 3.Interchange x and y in the equation found in step 2. This expresses f –1 as a function of x. Step 4.Find the domain of f –1. Remember, the domain of f –1 must be the same as the range of f. Check your work by verifying that: f –1 [f(x)] = xfor all x in the domain of f and f[f –1 (x)] = xfor all x in the domain of f –1 Finding the Inverse of a Function f

The graphs of y = f(x) and y = f – 1 (x) are symmetric with respect to the line y = x. x y 5–5 5 (1, 4) (4, 1) (2, –3) (–3, 2) (–5, –2) (–2, –5) y =x (a) (a, b) and (b, a) are symmetric with respect to the line y = x x y 5–5 5 y =x y =f(x) y =f(x) (b) f(x) = 2x – 1 f –1 (x) =  x +  Symmetry Property for the Graphs of f and f –