Lesson 5 Contents Example 1Two Rational Roots Example 2One Rational Root Example 3Irrational Roots Example 4Complex Roots Example 5Describe Roots

Quadratic Formula Video DzI DzI quadratic-formula/ quadratic-formula/

Example 5-1a Solve by using the Quadratic Formula. First, write the equation in the form and identify a, b, and c.

Example 5-1a Replace a with 1, b with –8, and c with –33. Simplify. Then, substitute these values into the Quadratic Formula. Quadratic Formula

Example 5-1a Answer: The solutions are 11 and –3. Write as two equations. or Simplify.

Example 5-1b Answer: 2, –15 Solve by using the Quadratic Formula.

Replace a with 1, b with –34, and c with 289. Identify a, b, and c. Then, substitute these values into the Quadratic Formula. Example 5-2a Solve by using the Quadratic Formula. Quadratic Formula Simplify.

Example 5-2a Answer: The solution is 17. Check A graph of the related function shows that there is one solution at

Solve by using the Quadratic Formula. Example 5-2b Answer: 11

Example 5-3a Solve by using the Quadratic Formula. Quadratic Formula Replace a with 1, b with –6, and c with 2. Simplify.

Example 5-3a Answer: The exact solutions are and The approximate solutions are 0.4 and 5.6. or

Example 5-3a CheckCheck these results by graphing the related quadratic function, Using the ZERO function of a graphing calculator, the approximate zeros of the related function are –2.9 and 0.9.

Example 5-3b Solve by using the Quadratic Formula. Answer: or approximately 0.7 and 4.3

Discriminant Video -discriminant/

Discriminant and Types of Roots Value of DiscriminantType and # of RootsExample Graph b 2 – 4ac > 0 and is a 2 real, rational roots perfect square b 2 – 4ac > 0 and is not a 2 real, irrational roots perfect square b 2 – 4ac = 0 1 real, rational root b 2 – 4ac < 0 2 complex roots

Example 5-5a Answer: The discriminant is 0, so there is one rational root. Find the value of the discriminant for. Then describe the number and type of roots for the equation.

Example 5-5a Answer: The discriminant is negative, so there are two complex roots. Find the value of the discriminant for. Then describe the number and type of roots for the equation.

Example 5-5a Answer: The discriminant is 80, which is not a perfect square. Therefore, there are two irrational roots. Find the value of the discriminant for. Then describe the number and type of roots for the equation.

Example 5-5a Answer: The discriminant is 81, which is a perfect square. Therefore, there are two rational roots. Find the value of the discriminant for. Then describe the number and type of roots for the equation.

Find the value of the discriminant for each quadratic equation. Then describe the number and type of roots for the equation. a. b. c. d. Example 5-5b Answer: 0 ; 1 rational root Answer: –24 ; 2 complex roots Answer: 5 ; 2 irrational roots Answer: 64 ; 2 rational roots

End of Lesson 5

Assignment P 318 #14, 18, 20, 24

Lesson 6 Contents Example 1Graph a Quadratic Function in Vertex Form Example 2Write y = x 2 + bx + c in Vertex Form Example 3Write y = ax 2 + bx + c in Vertex Form, a  1 Example 4Write an Equation Given Points

Example 6-1a Now use this information to draw the graph. Analyze Then draw its graph. h = 3 and k = 2 Answer: The vertex is at (h, k) or (3, 2) and the axis of symmetry is The graph has the same shape as the graph of but is translated 3 units right and 2 units up.

Example 6-1a Step 1 Plot the vertex, (3, 2). Step 3 Find and plot two points on one side of the axis of symmetry, such as (2, 3) and (1, 6). Step 4 Use symmetry to complete the graph. Step 2 Draw the axis of symmetry, (1, 6)(5, 6) (2, 3)(4, 3) (3, 2)

Example 6-1b Answer: Analyze Then draw its graph. The vertex is at (–2, – 4), and the axis of symmetry is The graph has the same shape as the graph of ; it is translated 2 units left and 4 units down.

Example 6-2a Write in vertex form. Then analyze the function. Notice that is not a perfect square. Balance this addition by subtracting 1. Complete the square by adding

Example 6-2a This function can be rewritten as So, and Answer: The vertex is at (–1, 3), and the axis of symmetry is Since the graph opens up and has the same shape as but is translated 1 unit left and 3 units up. Write as a perfect square.

Example 6-2b Write in vertex form. Then analyze the function. Answer: vertex: (–3, –4) ; axis of symmetry: opens up; the graph has the same shape as the graph of but it is translated 3 units left and 4 units down.

Example 6-3a Write in vertex form. Then analyze and graph the function. Original equation Group and factor, dividing by a. Complete the square by adding 1 inside the parentheses. Balance this addition by subtracting –2(1).

Example 6-3a Write as a perfect square. Now graph the function. Two points on the graph to the right of are (0, 2) and (0.5, –0.5). Use symmetry to complete the graph. Answer: The vertex form is So, andThe vertex is at (–1, 4) and the axis of symmetry is Since the graph opens down and is narrower than It is also translated 1 unit left and 4 units up.

Example 6-3a

Example 6-3b Write in vertex form. Then analyze and graph the function. Answer: vertex: (–1, 7) ; axis of symmetry: x = –1 ; opens down; the graph is narrower than the graph of y = x 2,and it is translated 1 unit left and 7 units up.

Example 6-4a Write an equation for the parabola whose vertex is at (1, 2) and passes through (3, 4). The vertex of the parabola is at (1, 2) so and Since (3, 4) is a point on the graph of the parabola, and Substitute these values into the vertex form of the equation and solve for a. Vertex form Substitute 1 for h, 2 for k, 3 for x, and 4 for y. Simplify.

Example 6-4a Subtract 2 from each side. Divide each side by 4. Answer:The equation of the parabola in vertex form is

Example 6-4a Check A graph ofverifies that the parabola passes through the point at (3, 4).

Answer: Example 6-4b Write an equation for the parabola whose vertex is at (2, 3) and passes through (–2, 1).

End of Lesson 6

Assignment P 326 #16, 20, 30, 32

Lesson 7 Contents Example 1Graph a Quadratic Inequality Example 2Solve ax 2 + bx + c  0 Example 3Solve ax 2 + bx + c  0 Example 4Write an Inequality Example 5Solve a Quadratic Inequality

Graphing Quadratic Inequalities inequality/how-to-solve-and-graph-quadratic- inequality.phphttp:// inequality/how-to-solve-and-graph-quadratic- inequality.php

Example 7-1a Graph Step 1 Graph the related quadratic equation, Since the inequality symbol is >, the parabola should be dashed.

Example 7-1a Graph Step 2 Test a point inside the parabola, such as (1, 2). So, (1, 2) is a solution of the inequality. (1, 2)

Example 7-1a Step 3 Shade the region inside the parabola. Graph

Example 7-1b Answer: Graph

Example 7-2a The solution consists of the x values for which the graph of the related quadratic function lies above the x -axis. Begin by finding the roots of the related equation. Solve by graphing. Related equation Factor. Solve each equation. Zero Product Property or

Example 7-2a Sketch the graph of the parabola that has x -intercepts at 3 and 1. The graph lies above the x -axis to the left of and to the right of Answer: The solution set is

Example 7-2b Solve by graphing. Answer:

Example 7-3a Solve by graphing. This inequality can be rewritten as The solution consists of the x -values for which the graph of the related quadratic equation lies on and above the x -axis. Begin by finding roots of the related equation. Related equation Use the Quadratic Formula. Replace a with –2, b with –6 and c with 1.

Example 7-3a Simplify and write as two equations. or Simplify. Sketch the graph of the parabola that has x -intercepts of –3.16 and The graph should open down since a < 0.

Example 7-3a The graph lies on and above the x -axis at and and between these two values. The solution set of the inequality is approximately Answer:

Example 7-3a Check Test one value of x less than – 3.16, one between –3.16 and 0.16, and one greater than 0.16 in the original inequality.

Example 7-3b Solve by by graphing. Answer:

Example 7-5a Solve algebraically. First, solve the related equation. Related quadratic equation Subtract 2 from each side. Factor. Solve each equation. Zero Product Property or

Example 7-5a Plot –2 and 1 on a number line. Use closed circles since these solutions are included. Notice that the number line is separated into 3 intervals.

Example 7-5a Test a value in each interval to see if it satisfies the original inequality.

Example 7-5a Answer: The solution set is This is shown on the number line below.

Example 7-5b Solve algebraically. Answer:

Assignment P 333 #14, 18, 26, 28

End of Lesson 7