1 Chapter 8 Quadratic Functions

2 Chapter Sections 8.1 – Solving Quadratic Equations by Completing the Square 8.2 – Solving Quadratic Equations by the Quadratic Formulas 8.3 – Quadratic Equations: Applications and Problem Solving 8.4 – Writing Equations in Quadratic Form 8.5 – Graphing Quadratic Functions 8.6 – Quadratic and Other Inequalities in One Variable Chapter 1 Outline

3 8.2 Solving Quadratic Equations by the Quadratic Formula1. Solve quadratic equations using the quadratic formula. 2. Use the discriminant to determine the number of real solutions that a quadratic equation has. 3. Find the x- and y-intercepts of a quadratic function. 4. Solve applications using the quadratic formula.

4 Using the Quadratic Formula To solve a quadratic equation in the standard form ax2 + bx + c = 0, where 𝑎≠0, use the quadratic formula: There once was a Negative Boy who couldn't decide to go to a Radical Party. Because the Boy was Square , he missed out on 4 Awesome Chicks and it was all over by 2 AM. Example 1: Solve x2 + 2x – 8 = 0. Solution a = 1, b = 2, and c = –8

5 Example 2: Solve by the quadratic FormulaSolution The equation is in the form ax2 + bx + c = 0, where a = 2, b = –7, and c = 3.

6 Example 3: Solve by the quadratic Formulaa = 1, b = –2, c = –11. The radicand is 0, which causes this equation to have only one solution.

7 Example 4: Solve by the quadratic FormulaNotice that the radicand is negative, which causes this equation to have 2 imaginary solutions.

8 Methods for Solving Quadratic EquationsDiscriminant: b2 – 4ac Solutions of a Quadratic Equation ax2 + bx + c = 0, a ≠ 0: b2 – 4ac > 0 → 2 distinct real solutions. The solutions will be rational if the discriminant is a perfect square and irrational otherwise. b2 – 4ac = 0 → 1 real solution. (Rational) b2 – 4ac < 0 → No real solution. (2 Complex solutions) Methods for Solving Quadratic Equations 1. Factoring Use when the quadratic equation can be easily factored. 2. Square root principle Use when the quadratic equation can be easily written in the form 3. Completing the square Rarely the best method, but important for future topics. 4. Quadratic formula Use when factoring is not easy, or possible, with integer coefficients.

9 Example 5: Use the discriminant to determine the number and type of solutions.1 real solution 2 real-number solutions. No real solutions (2 Imaginary or non-real complex solutions)

10 Discriminant Graphs of f(x) = ax2 + bx + cIf b2 – 4ac > 0, f(x) has two distinct x-intercepts. y x or y x

11 Discriminant Graphs of f(x) = ax2 + bx + c2. If b2 – 4ac = 0, f(x) has one single x-intercept. y x y x or

12 Discriminant Graphs of f(x) = ax2 + bx + c3. If b2 – 4ac < 0, f(x) has no x-intercept. y x y x or

13 Determine a Quadratic Equation Given Its SolutionsExample 6: Determine an equation that has the given solutions.

14 Example 7: Three times the square of a positive number decreased by twice the number is 21. Find the number. Example 9: Q94 A farmer has 400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. Find the dimensions of the field, if he wishes to enclose 15,000 𝑓𝑡 2 of area.

15 Example (If time permits)Find the x- and y-intercepts of y = x2 + x – 12, then graph. Solution Find the x-intercept by letting y = 0. 0 = x2 + x – 12 0 = (x + 4)(x – 3) x + 4 = or x – 3 = 0 x = –4 or x = 3 x-intercepts are (4, 0) and (3, 0) Find the y-intercept, letting x = 0. y = (0)2 + 0 – 12 = –12 y-intercept is (0, 12)

16 h = 15, v0 = 80, and h0 = 25 In physics, the general formula for describing the height of an object after it has been thrown upwards is where g represents the acceleration due to gravity, t is the time in flight, v0 is the initial velocity, and h0 is the initial height. For Earth, the acceleration due to gravity is  ft./sec.2 or 9.8 m/sec.2 a = 16.1, b = 80, c = 10. Answer Since the time cannot be negative , the motorcycle is in the air approximately 5.09 seconds.

17 Study Applications That Use Quadratic EquationsExample 6: Suppose the revenue from selling n cell phone, n ≤ 50, is R(n)=n(50-0.2n) where (50-0.2n) is the price per cell phone, in dollars. Find the revenue when 30 cell phones are sold. The revenue from selling 30 cell phones is $1,320.