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

2. Axis of Symmetry, y-intercept, and Vertex
A. Consider the quadratic function f(x) = 2 – 4x + x2. 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) = ax2 + bx + c
f(x) = 2 – 4x + x2
f(x) = 1x2 – 4x + 2
a = 1, b = –4, c = 2
The y-intercept is 2.
Example 2

3. Concept

4. Use a and b to find the equation of the axis of symmetry.
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.
Example 2

5. Axis of Symmetry, y-intercept, and Vertex
B. Consider the quadratic function f(x) = 2 – 4x + x2. 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
Example 2

6. Graph the vertex and y-intercept.
Axis of Symmetry, y-intercept, and Vertex
C. Consider the quadratic function f(x) = 2 – 4x + x2. 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.
Example 2

7. Axis of Symmetry, y-intercept, and Vertex
Answer:
Example 2

8. Concept

9. For this function, a = –1, b = 2, and c = 3.
Maximum or Minimum Values
A. Consider the function f(x) = –x2 + 2x + 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.
Example 3

10. The maximum value of this function is the y-coordinate of the vertex.
Maximum or Minimum Values
B. Consider the function f(x) = –x2 + 2x + 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.
Example 3

11. The domain is all real numbers.
Maximum or Minimum Values
C. Consider the function f(x) = –x2 + 2x + 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.
Example 3

12. Quadratic Functions Solving by: Graphing Pg:P13 Pb: 1 through 21 odd
LESSON 03
Quadratic Functions
Solving by:
Graphing
Pg:P13
Pb: 1 through 21 odd

13. Concept

14. Factor GCF
A. Solve 9y2 + 3y = 0
Example 2

15. 5a(a) - 5a(4) = 0 Factor the GCF.
Factor GCF
B. Solve 5a2 – 20a = 0
5a2 – 20a= 0
5a(a) - 5a(4) = 0 Factor the GCF.
5a(a - 4) =0 Distributive Property
5a = a - 4= 0 Zero Product Property Answer: a = 0, a = 4
Example 2

16. x 2 = (x)2; 9 = 32 First and last terms are perfect squares.
Perfect Squares and Differences of Squares
A. Solve x2 – 6x + 9 = 0
x 2 = (x)2; 9 = 32 First and last terms are perfect squares.
6x = 2(x)(3) Middle term equals 2ab.
x 2 - 6x + 9 is a perfect square trinomial.
x 2 - 6x + 9 = (x – 3)2 Factor using the pattern.
Example 3

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

18. y2 - 36= 0 Subtract 36 from both sides.
Perfect Squares and Differences of Squares
B. Solve y2 = 36.
y2 = 36 Original equation
y2 - 36= 0 Subtract 36 from both sides.
y2 - 62= 0 Write in the form a2 – b2 .
(y + 6) (y – 6) = 0 Factor the difference of squares.
Example 3

19. y + 6 = 0 or y – 6 = 0 Zero product property y = -6 or y = 6 Solve.
Perfect Squares and Differences of Squares
y + 6 = 0 or y – 6 = 0 Zero product property
y = -6 or y = 6 Solve.
Answer: y = -6 or y = 6
Example 3

20. Factor Trinomials
Solve x2 – 2x – 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
Example 4

21. x 2 - 2x – 15 = 0 Original equation.
Factor Trinomials
x 2 - 2x – 15 = 0 Original equation.
x 2 + mx + px – 15 = 0 Write the pattern.
x 2 + 3x + (–5)x – 15 = 0 m = 3, p = -5
(x 2 + 3x) + (–5x – 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
Example 4

22. x = -3 or x = 5 Solve Answer: x = - 3 or x = 5 Factor Trinomials
Example 4

23. 5x2 + 34x + 24 = 0 Original equation.
Factor Trinomials
B. Solve 5x2 + 34x + 24 = 0.
ac = 120 a = 5, c = 24
5x2 + 34x + 24 = 0 Original equation.
5x 2 + mx + px + 24 = 0 Write the pattern.
5x2 + 30x + 4x + 24 = 0 m = 30, p = 4
(5x x) + (4x + 24) = 0 Group terms with common factors.
5x(x + 6) + 4(x + 6) = 0 Factor the GCF from each grouping.
(5x+ 4)(x + 6) = 0 Distributive Property
Example 4

24. Factor Trinomials
Example 4

25. 9 – x 2 = 0 Original expression x 2 – 9 = 0 Multiply both sides by –1.
Solve Equations by Factoring
9 – x 2 = 0 Original expression
x 2 – 9 = 0 Multiply both sides by –1.
(x + 3)(x – 3) = 0 Difference of squares
x + 3 = 0 or x – 3 = 0 Zero Product Property
x = – x = 3 Solve.
Answer: The distance between 3 and – 3 is 3 – (–3) or 6 feet.
Example 5

26. Check 9 – x 2 = 0 9 – (3)2 = 0 or 9 – (–3)2 = 0 9 – 9 = 0 9 – 9 = 0
Solve Equations by Factoring
Check 9 – x 2 = 0
9 – (3)2 = 0 or 9 – (–3)2 = 0 ?
9 – 9 = 0 9 – 9 = 0 ?
0 = 0 0 = 0  
Example 5

27. Solve x 2 + 14x + 49 = 64 by using the Square Root Property.
Equation with Rational Roots
Solve x x + 49 = 64 by using the Square Root Property.
Original equation
Factor the perfect square trinomial.
Square Root Property
Subtract 7 from each side.
Example 1

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

29. Solve x 2 – 4x + 4 = 13 by using the Square Root Property.
Equation with Irrational Roots
Solve x 2 – 4x + 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.
Example 2

30. x 2 – 4x + 4 = 13 Original equation
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 – 4x + 4 = 13 Original equation
x 2 – 4x – 9 = 0 Subtract 13 from each side.
y = x 2 – 4x – 9 Related quadratic function
Example 2

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

32. Quadratic Functions Solving by: Factoring Pg: P13
LESSON 03
Quadratic Functions
Solving by:
Factoring
Pg: P13
Pb: 23 through 27 odd

33. Concept

34. Step 2 Square the result of Step 1. 62 = 36
Complete the Square
Find the value of c that makes x 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 = 36
Step 3 Add the result of Step 2 to x x + 36 x x.
Answer: The trinomial x2 + 12x + 36 can be written as (x + 6)2.
Example 3

35. Solve x2 + 4x – 12 = 0 by completing the square.
Solve an Equation by Completing the Square
Solve x2 + 4x – 12 = 0 by completing the square.
x2 + 4x – 12 = 0 Notice that x2 + 4x – 12 is not a perfect square.
x2 + 4x = 12 Rewrite so the left side is of the form x2 + bx.
x2 + 4x + 4 = add 4 to each side.
(x + 2)2 = 16 Write the left side as a perfect square by factoring.
Example 4

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

37. Solve 3x2 – 2x – 1 = 0 by completing the square.
Equation with a ≠ 1
Solve 3x2 – 2x – 1 = 0 by completing the square.
3x2 – 2x – 1 = 0 Notice that 3x2 – 2x – 1 is not a perfect square.
Divide by the coefficient of the quadratic term, 3.
Add to each side.
Example 5

38. Equation with a ≠ 1
Write the left side as a perfect square by factoring. Simplify the right side.
Square Root Property
Example 5

39. or Write as two equations.
Equation with a ≠ 1
or Write as two equations.
x = 1 Solve each equation.
Answer:
Example 5

40. Solve x 2 + 4x + 11 = 0 by completing the square.
Equation with Imaginary Solutions
Solve x 2 + 4x + 11 = 0 by completing the square.
Notice that x 2 + 4x + 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
Example 6

41. Subtract 2 from each side.
Equation with Imaginary Solutions
Subtract 2 from each side.
Example 6

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

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

44. Concept

45. Solve x2 – 8x = 33 by using the Quadratic Formula.
Two Rational Roots
Solve x2 – 8x = 33 by using the Quadratic Formula.
First, write the equation in the form ax2 + bx + c = 0 and identify a, b, and c.
ax2 + bx + c = 0
x2 – 8x = 33 1x2 – 8x – 33 = 0
Then, substitute these values into the Quadratic Formula.
Quadratic Formula
Example 1

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

47. or Write as two equations.
Two Rational Roots
or Write as two equations.
x = x = –3 Simplify.
Answer: The solutions are 11 and –3.
Example 1

48. Solve x2 – 34x + 289 = 0 by using the Quadratic Formula.
One Rational Root
Solve x2 – 34x = 0 by using the Quadratic Formula.
Identify a, b, and c. Then, substitute these values into the Quadratic Formula.
Quadratic Formula
Replace a with 1, b with –34, and c with 289.
Simplify.
Example 2

49. Answer: The solution is 17.
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
Example 2

50. Solve x2 – 6x + 2 = 0 by using the Quadratic Formula.
Irrational Roots
Solve x2 – 6x + 2 = 0 by using the Quadratic Formula.
Quadratic Formula
Replace a with 1, b with –6, and c with 2.
Simplify. or
Example 3

51. Irrational Roots
Answer:
Check Check these results by graphing the related quadratic function, y = x2 – 6x + 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
Example 3

52. Solve x2 + 13 = 6x by using the Quadratic Formula.
Complex Roots
Solve x = 6x by using the Quadratic Formula.
Quadratic Formula
Replace a with 1, b with –6, and c with 13.
Simplify.
Simplify.
Example 4

53. Answer: The solutions are the complex numbers 3 + 2i and 3 – 2i.
Complex Roots
Answer: The solutions are the complex numbers 3 + 2i and 3 – 2i.
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
Example 4

54. x2 + 13 = 6x Original equation
Complex Roots
Check To check complex solutions, you must substitute them into the original equation. The check for 3 + 2i is shown below.
x = 6x Original equation
(3 + 2i) = 6(3 + 2i) x = (3 + 2i) ?
9 + 12i + 4i = i Square of a sum; Distributive Property ?
i – 4 = i Simplify. ?
i = i 
Example 4

55. Concept

56. b2 – 4ac = (3)2 – 4(1)(5) Substitution = 9 – 20 Simplify.
Describe Roots
A. Find the value of the discriminant for x2 + 3x + 5 = 0. Then describe the number and type of roots for the equation.
a = 1, b = 3, c = 5
b2 – 4ac = (3)2 – 4(1)(5) Substitution
= 9 – 20 Simplify.
= –11 Subtract.
Answer: The discriminant is negative, so there are two complex roots.
Example 5

57. b2 – 4ac = (–11)2 – 4(1)(10) Substitution = 121 – 40 Simplify.
Describe Roots
B. Find the value of the discriminant for x2 – 11x + 10 = 0. Then describe the number and type of roots for the equation.
a = 1, b = –11, c = 10
b2 – 4ac = (–11)2 – 4(1)(10) Substitution
= 121 – 40 Simplify.
= 81 Subtract.
Answer: The discriminant is 81, so there are two rational roots.
Example 5

58. Concept

59. Quadratic Functions Solving by: Quadratic equation Pg: P13
LESSON 03
Quadratic Functions
Solving by:
Quadratic equation
Pg: P13
Pb: 35 through 39 odd