Algebra lesson 4.2 zeroes of quadratic functions
- 1. QUADRATIC FUNCTIONS MIDTERM EXAM: FEB 1, W, 10-11:30 AM BAR AND DINING LAB (Bring Calculator) Lesson 4.2 Real Zeroes Factoring Quadratic Formula
- 2. Zeroes • The values of x so that f(x) = 0. • Real (as opposed to imaginary) zeroes are also the x-intercepts of the function.
- 3. Zeroes of Quadratic Functions CASE 1: FACTORABLE QEs • The Zero Product Theorem For a product to be equal to zero, at least one of its factors must be equal to zero. i.e. If abcd = 0, then either a=0 or b=0 or c=0 or d=0. • To solve (x+2)(x-3)(2x+1)=0, find values of x that will make each factor equal to zero. • Hence, the solutions are x = -2, 3 and -1/2.
- 4. Zeroes of Quadratic Functions Example 1: Find the x-intercept/s of f(x) = x2 – 9. Solution: 0 = x2 – 9 0 = (x – 3)(x + 3) Hence, x = 3 and -3
- 5. Answers to the SW: 1. 0 = 4x2 – 9 0 = (2x + 3)(2x – 3) Solution: x = -3/2 and 3/2 2. 0 = x2 - 5x – 24 0 = (x – 8)(x + 3) Solution: x = 8, -3 3. 20 = 9x - x2 0 = - x2 + 9x – 20 = (-x + 4)(x – 5) Solution: x = 4, 5 4. 0 = 3x2 + 4x – 4 Solution: x = 2/3, -2 5. 3 = 4x2 + 4x Solution: x = -3/2, 1/2
- 6. Quadratic Formula • Not all quadratic equations are factorable but that does not mean that it does not have zeroes. • The following formula solves for the value of x that satisfies the equation 0 = ax2 + bx + c: • Notice that this is actually two formulas, one using + and the other using -
- 7. Quadratic Formula Example: Find the x-intercepts of f(x)=x2 – 3x – 1 Solution: a = 1, b = -3, c = -1 -(-3) ± (-3)2 - 4(1)(-1) x= 2(1) 3± 9 + 4 x= 3+ 13 3- 13 2 x= and 2 2 3± 13 x= 2
- 8. Seatwork 1. Solve for x: a) 12 = x2 + x c) 2x2 + 6 = 7x b) 0 = 2x2 – 2x – 3 d) 0 = -x2 + 3x + 5 2. Given the function f(x) = x2 + 8x + 10, do the ff: a) Find the vertex, y-intercept and x-intercepts b) Graph f(x) showing the points in (a). c) Identify the domain, range, axis of symmetry and interval/s where f(x) is positive.