LESSON 03 Quadratic Functions Solving by Graphing Factoring

LESSON 03 Quadratic Functions Solving by: • Graphing • • • Factoring Completing the square Quadratic equation

Axis of Symmetry, y-intercept, and Vertex A. Consider the quadratic function f(x) = 2 – 4 x + x 2. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex. Begin by rearranging the terms of the function so that the quadratic term is first, the linear term is second and the constant term is last. Then identify a, b, and c. f(x) = ax 2 + bx + c f(x) = 2 – 4 x + x 2 The y-intercept is 2. f(x) = 1 x 2 – 4 x + 2 a = 1, b = – 4, c = 2

Axis of Symmetry, y-intercept, and Vertex Use a and b to find the equation of the axis of symmetry. Equation of the axis of symmetry a = 1, b = – 4 x =2 Simplify. Answer: The y-intercept is 2. The equation of the axis of symmetry is x = 2. Therefore, the x-coordinate of the vertex is 2.

Axis of Symmetry, y-intercept, and Vertex B. Consider the quadratic function f(x) = 2 – 4 x + x 2. Make a table of values that includes the vertex. Choose some values for x that are less than 2 and some that are greater than 2. This ensures that points on each side of the axis of symmetry are graphed. Answer: Vertex

Axis of Symmetry, y-intercept, and Vertex C. Consider the quadratic function f(x) = 2 – 4 x + x 2. Use the information from parts A and B to graph the function. Graph the vertex and y-intercept. Then graph the points from your table, connecting them with a smooth curve. As a check, draw the axis of symmetry, x = 2, as a dashed line. The graph of the function should be symmetrical about this line.

Axis of Symmetry, y-intercept, and Vertex Answer:

Maximum or Minimum Values A. Consider the function f(x) = –x 2 + 2 x + 3. Determine whether the function has a maximum or a minimum value. For this function, a = – 1, b = 2, and c = 3. Answer: Since a < 0, the graph opens down and the function has a maximum value.

Maximum or Minimum Values B. Consider the function f(x) = –x 2 + 2 x + 3. State the maximum or minimum value of the function. The maximum value of this function is the y-coordinate of the vertex. Find the y-coordinate of the vertex by evaluating the function for x = 1. Answer: The maximum value of the function is 4.

Maximum or Minimum Values C. Consider the function f(x) = –x 2 + 2 x + 3. State the domain and range of the function. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value.

LESSON 03 Quadratic Functions Solving by: • Graphing Pg: P 13 Pb: 1 through 21 odd

Factor GCF A. Solve 9 y 2 + 3 y = 0

Factor GCF B. Solve 5 a 2 – 20 a = 0 5 a 2 – 20 a= 0 5 a(a) - 5 a(4) = 0 5 a(a - 4) =0 5 a = 0 a - 4= 0 Answer: a = 0, a = 4 Factor the GCF. Distributive Property Zero Product Property

Perfect Squares and Differences of Squares A. Solve x 2 – 6 x + 9 = 0 x 2 = (x)2; 9 = 32 First and last terms are perfect squares. 6 x = 2(x)(3) Middle term equals 2 ab. x 2 - 6 x + 9 is a perfect square trinomial. x 2 - 6 x + 9 = (x – 3)2 Factor using the pattern.

Perfect Squares and Differences of Squares x 2 - 6 x + 9 = 0 Original equation (x – 3)2 = 0 Substitution. x – 3 = 0 Take the square root of each side. Answer: x = 3

Perfect Squares and Differences of Squares B. Solve y 2 = 36 Original equation y 2 - 36= 0 Subtract 36 from both sides. y 2 - 62= 0 Write in the form a 2 – b 2. (y + 6) (y – 6) = 0 Factor the difference of squares.

Perfect Squares and Differences of Squares y + 6 = 0 or y – 6 = 0 Zero product property y = -6 or y = 6 Answer: y = -6 or y = 6 Solve.

Factor Trinomials A. Solve x 2 – 2 x – 15 = 0 Find two values, m and p, such that their product equals ac and their sum equals b. b = -2 ac = – 15 a = 1, c = – 15 Factors of -15 Sum -1, 15 14 1, -15 -14 -3, 5 2 -5, 3 -2

Factor Trinomials x 2 - 2 x – 15 = 0 Original equation. x 2 + mx + px – 15 = 0 Write the pattern. x 2 + 3 x + (– 5)x – 15 = 0 m = 3, p = -5 (x 2 + 3 x) + (– 5 x – 15) = 0 Group terms with common factors. x(x + 3) – 5(x + 3) = 0 Factor the GCF from each grouping. (x + 3)(x – 5) = 0 Distributive Property x + 3 = 0 or x – 5 = 0 Zero product property

Factor Trinomials x = -3 or x = 5 Answer: x = - 3 or x = 5 Solve

Factor Trinomials B. Solve 5 x 2 + 34 x + 24 = 0. ac = 120 a = 5, c = 24 5 x 2 + 34 x + 24 = 0 Original equation. 5 x 2 + mx + px + 24 = 0 Write the pattern. 5 x 2 + 30 x + 4 x + 24 = 0 m = 30, p = 4 (5 x 2 + 30 x) + (4 x + 24) = 0 Group terms with common factors. 5 x(x + 6) + 4(x + 6) = 0 Factor the GCF from each grouping. (5 x+ 4)(x + 6) = 0 Distributive Property

Factor Trinomials

Solve Equations by Factoring 9 – x 2 = 0 Original expression x 2 – 9 = 0 Multiply both sides by – 1. (x + 3)(x – 3) = 0 x + 3 = 0 or x – 3 = 0 x = – 3 x = 3 Difference of squares Zero Product Property Solve. Answer: The distance between 3 and – 3 is 3 – (– 3) or 6 feet.

Solve Equations by Factoring Check 9 – x 2 = 0 2 ? 9 – (3) = 0 ? 9 – 9 = 0 0 = 0 or 2 ? 9 – (– 3) = 0 ? 9 – 9 = 0 0 =0

Equation with Rational Roots Solve x 2 + 14 x + 49 = 64 by using the Square Root Property. Original equation Factor the perfect square trinomial. Square Root Property Subtract 7 from each side.

Equation with Rational Roots x = – 7 + 8 or x = – 7 – 8 Write as two equations. x = 1 Solve each equation. x = – 15 Answer: The solution set is {– 15, 1}. Check: Substitute both values into the original equation. x 2 + 14 x + 49 = 64 2 ? 1 + 14(1) + 49= 64 ? 1 + 14 + 49 = 64 64 = 64 x 2 + 14 x + 49 = 64 2 ? (– 15) + 14(– 15) + 49 = 64 ? 225 + (– 210) + 49 = 64 64 = 64

Equation with Irrational Roots Solve x 2 – 4 x + 4 = 13 by using the Square Root Property. Original equation Factor the perfect square trinomial. Square Root Property Add 2 to each side. Write as two equations. Use a calculator.

Equation with Irrational Roots Answer: The exact solutions of this equation are The approximate solutions are 5. 61 and – 1. 61. Check these results by finding and graphing the related quadratic function. x 2 – 4 x + 4 =13 Original equation x 2 – 4 x – 9 =0 Subtract 13 from each side. y = x 2 – 4 x – 9 Related quadratic function

LESSON 03 Quadratic Functions Solving by: • Graphing • Factoring • Completing the square • Quadratic equation

LESSON 03 Quadratic Functions Solving by: • Factoring Pg: P 13 Pb: 23 through 27 odd

Complete the Square Find the value of c that makes x 2 + 12 x + c a perfect square. Then write the trinomial as a perfect square. Step 1 Find one half of 12. Step 2 Square the result of Step 1. 62 = 36 Step 3 Add the result of Step 2 to x 2 + 12 x + 36 Answer: The trinomial x 2 + 12 x + 36 can be written as (x + 6)2.

Solve an Equation by Completing the Square Solve x 2 + 4 x – 12 = 0 by completing the square. x 2 + 4 x – 12 = 0 Notice that x 2 + 4 x – 12 is not a perfect square. x 2 + 4 x = 12 Rewrite so the left side is of the form x 2 + bx. x 2 + 4 x + 4 = 12 + 4 add 4 to each side. (x + 2)2 = 16 Write the left side as a perfect square by factoring.

Solve an Equation by Completing the Square x + 2 = ± 4 x = – 2 ± 4 Square Root Property Subtract 2 from each side. x = – 2 + 4 or x = – 2 – 4 Write as two equations. x = 2 Solve each equation. x = – 6 Answer: The solution set is {– 6, 2}.

Equation with a ≠ 1 Solve 3 x 2 – 2 x – 1 = 0 by completing the square. 3 x 2 – 2 x – 1 = 0 Notice that 3 x 2 – 2 x – 1 is not a perfect square. Divide by the coefficient of the quadratic term, 3. Add to each side.

Equation with a ≠ 1 Write the left side as a perfect square by factoring. Simplify the right side. Square Root Property

Equation with a ≠ 1 or x = 1 Answer: Write as two equations. Solve each equation.

Equation with Imaginary Solutions Solve x 2 + 4 x + 11 = 0 by completing the square. Notice that x 2 + 4 x + 11 is not a perfect square. Rewrite so the left side is of the form x 2 + bx. Since , add 4 to each side. Write the left side as a perfect square. Square Root Property

Equation with Imaginary Solutions Subtract 2 from each side.

LESSON 03 Quadratic Functions Solving by: • Graphing • Factoring • Completing the square • Quadratic equation

LESSON 03 Quadratic Functions Solving by: • Completing the square Pg: 13 Pb: 29 through 33 odd

Two Rational Roots Solve x 2 – 8 x = 33 by using the Quadratic Formula. First, write the equation in the form ax 2 + bx + c = 0 and identify a, b, and c. ax 2 + bx + c = 0 x 2 – 8 x = 33 1 x 2 – 8 x – 33 = 0 Then, substitute these values into the Quadratic Formula

Two Rational Roots Replace a with 1, b with – 8, and c with – 33. Simplify.

Two Rational Roots or x = 11 x = – 3 Write as two equations. Simplify. Answer: The solutions are 11 and – 3.

One Rational Root Solve x 2 – 34 x + 289 = 0 by using the Quadratic Formula. Identify a, b, and c. Then, substitute these values into the Quadratic Formula Replace a with 1, b with – 34, and c with 289. Simplify.

One Rational Root Answer: The solution is 17. Check A graph of the related function shows that there is one solution at x = 17. [– 5, 25] scl: 1 by [– 5, 15] scl: 1

Irrational Roots Solve x 2 – 6 x + 2 = 0 by using the Quadratic Formula Replace a with 1, b with – 6, and c with 2. Simplify. or or

Irrational Roots Answer: Check these results by graphing the related quadratic function, y = x 2 – 6 x + 2. Using the ZERO function of a graphing calculator, the approximate zeros of the related function are 0. 4 and 5. 6. [– 10, 10] scl: 1 by [– 10, 10] scl: 1

Complex Roots Solve x 2 + 13 = 6 x by using the Quadratic Formula Replace a with 1, b with – 6, and c with 13. Simplify.

Complex Roots Answer: The solutions are the complex numbers 3 + 2 i and 3 – 2 i. A graph of the related function shows that the solutions are complex, but it cannot help you find them. [– 5, 15] scl: 1 by [– 5, 15] scl: 1

Complex Roots Check To check complex solutions, you must substitute them into the original equation. The check for 3 + 2 i is shown below. x 2 + 13 = 6 x 2 ? (3 + 2 i) + 13 = 6(3 + 2 i) 9 + 12 i + 4 i + 13 =18 + 12 i ? 22 + 12 i – 4 = 18 + 12 i 18 + 12 i = 18 + 12 i Original equation x = (3 + 2 i) Square of a sum; Distributive Property Simplify.

Describe Roots A. Find the value of the discriminant for x 2 + 3 x + 5 = 0. Then describe the number and type of roots for the equation. a = 1, b = 3, c = 5 b 2 – 4 ac = (3)2 – 4(1)(5) Substitution = 9 – 20 Simplify. = – 11 Subtract. Answer: The discriminant is negative, so there are two complex roots.

Describe Roots B. Find the value of the discriminant for x 2 – 11 x + 10 = 0. Then describe the number and type of roots for the equation. a = 1, b = – 11, c = 10 b 2 – 4 ac = (– 11)2 – 4(1)(10) Substitution = 121 – 40 Simplify. = 81 Subtract. Answer: The discriminant is 81, so there are two rational roots.

LESSON 03 Quadratic Functions Solving by: • Quadratic equation Pg: P 13 Pb: 35 through 39 odd