Chapter 4: Rational Power, and Root Functions 4.1 Rational Functions and Graphs (I) 4.2 Rational Functions and Graphs (II) 4.3 Rational Equations, Inequalities, Models, and Applications 4.4 Functions Defined by Powers and Roots 4.5 Equations, Inequalities, and Applications Involving Root Functions

4.1 Rational Functions and Graphs

4.1 The Reciprocal Function

4.1 The Rational Function f(x) = 1/x2

Rational Functions Given: 𝒚=𝒂 𝟏 𝒙−𝒉 +𝒌 𝒚=𝒂 𝟏 𝒙−𝒉 𝟐 +𝒌 How to find: 𝒚=𝒂 𝟏 𝒙−𝒉 +𝒌 𝒚=𝒂 𝟏 𝒙−𝒉 𝟐 +𝒌 How to find: Restrictions: Domain,Vertical Asymptote or Hole Intercepts: both x-intercept and y-intercept Horizontal or Oblique (Slant) Asympotote Rough Sketch

Rational Functions and transformations w/a, h and k Graph using a, h and k. 1. 𝑓 𝑥 = −2 𝑥+1 +2 2. 𝑔 𝑥 = 1 𝑥−2 2 −3 Change to a, h and k form: 3. 𝑓 𝑥 = 2𝑥−3 𝑥−1

Rational Functions and transformations w/a, h and k Graph using a, h and k. 1. 𝑓 𝑥 = −2 𝑥+1 +2 2. 𝑔 𝑥 = 1 𝑥−2 2 −3 Change to a, h and k form: 3. 𝑓 𝑥 = 2𝑥−3 𝑥−1 𝑓 𝑥 =2− 1 𝑥−1 changes to using synthetic or long division

Find the Restriction: Domain, Vertical Asymptote or Hole? 1. 𝑓 𝑥 = 𝑥+1 2 𝑥 2 +5𝑥−3 2. 𝑔 𝑥 = 2𝑥+1 𝑥−3 3. ℎ 𝑥 = 𝑥 2 +1 𝑥−2 4. 𝑘 𝑥 = 𝑥 2 −4 𝑥−2 Finding Restrictions – set denominator = 0 Domain: All Real numbers except x≠𝑡ℎ𝑒 𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑠 Hole: if you can simplify the rational equation Vertical Asymptote: Not Simplify then the restrictions are the VAs.

Finding the Horizontal Asymptotes or Oblique Asymptote: 1. 𝑓 𝑥 = 𝑥+1 2 𝑥 2 +5𝑥−3 2. 𝑔 𝑥 = 2𝑥+1 𝑥−3 3. ℎ 𝑥 = 𝑥 2 +1 𝑥−2 4. 𝑘 𝑥 = 𝑥 2 −4 𝑥−2 Case 1: Numerator Degree < Denominator Degree HA: y = 0 and no Slant (Oblique Asymptote) Case 2: Numerator Degree = Denominator Degree HA: y = Coeff of highest degree term/coeff of highest degree term and no Slant (Oblique Asymptote) Case 3: Numerator Degree > Denominator Degree No HA, but if the ND is only 1 degree bigger than the the DD there is a Slant (Oblique Asymptote)

The intercepts: x-intercept(s) and y-intercept 1. 𝑓 𝑥 = 𝑥+1 2 𝑥 2 +5𝑥−3 2. 𝑔 𝑥 = 2𝑥+1 𝑥−3 3. ℎ 𝑥 = 𝑥 2 +1 𝑥−2 4. 𝑘 𝑥 = 𝑥 2 −4 𝑥−2 How do you find them? x-intercepts: set numerator = 0 Y-intercept: substitute 0 in for x.

Comprehensive Graph Criteria for a Rational Function A comprehensive graph of a rational function will exhibits these features: all intercepts, both x and y; location of all asymptotes: vertical, horizontal, and/or oblique; the point at which the graph intersects its non-vertical asymptote (if there is such a point); enough of the graph to exhibit the correct end behavior (i.e. behavior as the graph approaches its nonvertical asymptote).

Graphing a Rational Function To sketch its graph, follow these steps: Find the domain and all vertical asymptotes. Find any horizontal or oblique asymptotes. Find the y-intercepts. Find the x-intercepts. Determine whether the graph will intersect its non-vertical asymptote by solving f (x) = k where y = k is the horizontal asymptote, or f (x) = mx + b where y = mx + b is the equation of the oblique asymptote. Plot a few selected points, as necessary. Choose an x- value between the vertical asymptotes and x-intercepts. Complete the sketch.

Example 1 Graphed Complete the sketch. The graph approaches its asymptotes as the points become farther away from the origin.

Example 2 Graphed

Example 3 Graphed

Example 4 Graphed

Chapter 3: Polynomial Functions 3.1 Complex Numbers 3.2 Quadratic Functions and Graphs 3.3 Quadratic Equations and Inequalities 3.4 Further Applications of Quadratic Functions and Models 3.5 Higher-Degree Polynomial Functions and Graphs 3.6 Topics in the Theory of Polynomial Functions (I) 3.7 Topics in the Theory of Polynomial Functions (II) 3.8 Polynomial Equations and Inequalities; Further Applications and Models

Solving Polynomial Equations: Zero- Product Property

Solving Polynomial Equations: Zero- Product Property

Solving an Equation Quadratic in Form Example Solve x4 − 6x2 − 40 = 0 analytically. Find all complex solutions.

Solving an Equation Quadratic in Form Example Solve x4 − 6x2 − 40 = 0 analytically. Find all complex solutions.

Solving a Polynomial Equation Example Show that 2 is a solution of x3 + 3x2 − 11x + 2 = 0, and then find all solutions of this equation.

Solving a Polynomial Equation Example Show that 2 is a solution of x3 + 3x2 − 11x + 2 = 0, and then find all solutions of this equation. Solution Use synthetic division.

Solving a Polynomial Equation P(x) = (x − 2) (x2 + 5x − 1) To find the other zeros of P, solve x2 + 5x − 1 = 0. Use the quadratic formula, with a = 1, b = 5, and c = −1,

Solving a Polynomial Equation P(x) = (x − 2) (x2 + 5x − 1) To find the other zeros of P, solve x2 + 5x − 1 = 0. Using the quadratic formula, with a = 1, b = 5, and c = −1,

Using Graphical Methods to Solve a Polynomial Equation Example Let P(x) = 2.45x3 − 3.14x2 − 6.99x + 2.58. Use the graph of P to solve P(x) = 0, P(x) > 0, and P(x) < 0.

Using Graphical Methods to Solve a Polynomial Equation Example Let P(x) = 2.45x3 − 3.14x2 − 6.99x + 2.58. Use the graph of P to solve P(x) = 0, P(x) > 0, and P(x) < 0. Solution

Applications and Polynomial Models Example A box with an open top is to be constructed from a rectangular 12-inch by 20-inch piece of cardboard by cutting equal-sized squares from each corner and folding up the sides. If x represents the length of the side of each square, determine a function V that describes the volume of the box in terms of x. Determine the value of x for which the volume of the box is maximized. What is this volume?

Applications and Polynomial Models Solution Volume = length  width  height V(x) = (20 − 2x)(12 − 2x)(x) = 4x3 − 64x2 + 240x where 0 < x < 6 Use the graph of V to find the local maximum point.