1. Objective: To find the zeros of a quadratic function and solve a quadratic equation

2.  Quadratic Equation: ◦ A second-degree equation in one variable. ◦ Has the form ax 2 + bx + c = 0  Quadratic Function: ◦ A second-degree function ◦ Has the form f(x) = ax 2 + bx + c  Zero of a Function: ◦ The point or points where the graph of the function crosses the x-axis ◦ Can be found by determining where f(x) = 0

3.  Zero-Product Property: ◦ If a multiplication problem has a product of zero, the one of the factors MUST equal zero  Solution of a Quadratic Equation: ◦ The value(s) of x which satisfy the equation ◦ The solutions of any quadratic equation are the same as the zeros of the related quadratic function  When finding the zeros of a function or the solutions to an equation, the number of zeros or solutions will be equal to the degree of the function/equation.

4.  Graphing  Factoring  Using square roots  Completing the square  Using the Quadratic Formula

5.  Use the graph below to determine the zeros of the function f(x) = x 2 – 3x - 10 The zeros of the function are -2 and 5

6.  Determine the zeros f(x) = x 2 – 3x – 10 by factoring ◦ Set the function equal to zero, then factor ◦ x 2 – 3x – 10 = 0 ◦ (x – 5)(x + 2) = 0 ◦ Use the zero-product property to determine the zeros x – 5 = 0 x = 5 orx + 2 = 0 x = -2 The zeros of the function are -2 and 5

7.  Determine the zeros of f(x) = 2x 2 + 3x – 9 by factoring  2x 2 + 3x – 9 = 0  (2x - 3)(x + 3) = 0 2x – 3 = 0 2x = 3 orx + 3 = 0 x = -3 The zeros of the function are -3 and 3/2

8.  Determine the zeros of f(x) = 9x 2 – 30x + 25 by factoring  9x 2 - 30x + 25 = 0  (3x - 5) 2 = 0 (3x – 5) 2 = 0 3x = 5 This function has a double zero of 5/3 The graph of the function touches the x-axis at the double zero, rather than crossing it. This is called a point of tangency. 3x – 5 = 0

9.  Determine the zeros of f(x) = 16x 2 – 49 by factoring  16x 2 – 49 = 0  (4x + 7)(4x – 7) = 0 4x + 7 = 0 4x = -7 or4x - 7 = 0 4x = 7 The zeros of the function are -7/4 and 7/4

10.  Determine the zeros of f(x) = 16x 2 – 49 by using square roots ◦ Because this function has no first-degree term, it can be solved by using square roots  16x 2 – 49 = 0 The zeros of the function are -7/4 and 7/4

11.  Determine the zeros of f(x) = x 2 + 4x – 2 ◦ This function cannot be factored ◦ This function cannot be solved by using square roots ◦ Graphing the function will show that the zeros are non-terminating, non-repeating decimals  The exact values of the zeros of this function can be found by using a method called Completing the Square  Completing the square is a process which turns the function into a perfect square trinomial

12.  To complete the square, use the following steps ◦ Set the function equal to zero ◦ Move the constant to the other side of the equation ◦ If the leading coefficient is not 1, divide each term by the leading coefficient ◦ Determine the number needed to “complete the square” by taking ½ of the first-degree term’s coefficient, squaring it, and adding it to both sides of the equation ◦ Factor the resulting perfect square trinomial ◦ Determine the zeros by using square roots

13.  Determine the zeros of f(x) = x 2 + 4x – 2  x 2 + 4x – 2 = 0  x 2 + 4x = 2  (x + 2) 2 = 6 Take ½ of 4 which is 2, square it to get 4, then add 4 to both sides + 4 The zeros of this function are and

14.  Determine the zeros of f(x) = x 2 - 6x + 4  x 2 - 6x + 4 = 0  x 2 - 6x = -4  (x - 3) 2 = 5 Take ½ of -6 which is -3, square it to get 9, then add 9 to both sides + 9 The zeros of this function are and

15.  Determine the zeros of f(x) = 2x 2 - 5x + 1  2x 2 - 5x + 1 = 0  2x 2 – 5x = -1   Take ½ of -5/2 which is -5/4, square it to get 25/16, then add 25/16 to both sides The zeros of this function are and As the leading coefficient is not 1, divide both sides by 2

16.  Solving a quadratic equation is very similar to finding the zeros of a quadratic function ◦ Get the equation into standard form ◦ Solve by graphing, factoring, using square roots or completing the square

17.  Solve:  3x 2 – 2x = 1  3x 2 – 2x – 1 = 0  (3x + 1)(x – 1) = 0  3x + 1 = 0orx – 1 = 0  3x = -1x = 1  x = -1/3x = 1  The solution set to this equation is {-1/3, 1}

18.  Solve:  4x 2 – 6x = 0  2x(2x – 3) = 0  2x = 0or2x – 3 = 0  x = 02x = 3  x = 0x = 3/2  The solution set to this equation is {0, 3/2}

19.  Solve:  3(x – 2) 2 – 6 = 0  3(x – 2) 2 = 6  (x – 2) 2 = 2  x – 2 =  x =  The solution set to this equation is { }

20.  The graph of a quadratic function is a parabola and can be graphed using its zeros. ◦ Determine the zeros using any method. ◦ Plot the zeros on the graph. ◦ The vertex of the parabola is halfway between the zeros.  Determine the midpoint of the zeros. This is the x- value of the vertex.  Substitute into the function to determine the y-value of the vertex. ◦ The y-intercept of the parabola can be found by substituting 0 for x and solving for y. ◦ Use the points to sketch the graph of the parabola.

21.  Graph f(x) = x 2 – 2x – 8  Factor: f(x) = (x – 4)(x + 2)  The zeros of the function are 4 and -2  The midpoint of the zeros is 1  Plug 1 in for x and get -9, so the vertex is (1, -9)  Plug zero in for x and get -8, so the y-intercept is (0, -8)  Sketch