Answer Key Name: ___________________________________
Date: _________________
SHIFTING PARABOLAS AND COMPLETING THE SQUARE ALGEBRA 2 WITH TRIGONOMETRY Parabolas, and graphs more generally, can be moved horizontally and vertically by simple manipulations of their equations. This is known as shifting or translating a graph. These manipulations are illustrated in the first exercise. Exercise #1: The function y x 2 is shown already graphed on the grid below. For parts (a) and (b), graph the function given and describe how the graph of y x 2 has been shifted in order to produce the new graph. y x2
(a) y x 4 2 2
y
y x 4 2 2
The graph of y x 2 has been shifted 4 units to the right and 2 units up. y x 3 5 2
x
(b) y x 3 5 2
The graph of y x 2 has been shifted 3 units to the left and 5 units down.
Exercise #2: If the parabola y x 2 were shifted 6 units left and 2 units down, its resulting equation would be which of the following? Verify by graphing your answer and seeing if its turning point is at 6, 2 . (1) y x 6 2
(3) y x 6 2
(2) y x 6 2
(4) y x 6 2
2
2
2
(2)
2
Exercise #3: Which of the following represents the turning point of the function f x x 8 4 ? 2
(1) 8, 4
(3) 8, 4
(2) 8, 4
(4) 8, 4
Since this graph represents y x 2 after a shift of 8 units right and 4 units down, our turning point is at 8, 4
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #3 – QUADRATIC FUNCTIONS AND THEIR ALGEBRA – LESSON #12 eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
(4)
Every parabola can be placed in what is known as its vertex form. For simple parabolas that are shifts of the function y x 2 , this will always take on the following form. THE VERTEX FORM OF A PARABOLA y x h k 2
where h, k is the vertex or turning point of the parabola.
It is possible to place every quadratic function of the form y x 2 bx c in the above form by a process known as completing the square. The following exercise will illustrate the technique. Exercise #4: Consider the quadratic y x 2 8x 7 . (a) Write this quadratic function in the form y x h k by completing the square. 2
Always square b 2
y x2 8 x 4 4 7 2
2
(b) Identify the turning point of the parabola from your answer in part (a). Verify your answer using tables on your calculator.
y x 2 8 x 16 16 7
y x 2 has been shifted 4 left and 9 down.
y x 4 9
Thus, the new turning point is at 4, 9.
2
Although very algorithmic in nature, the technique of completing the square arises often in mathematics and is extremely useful in determining the turning point of a parabola. Exercise #5: Which of the following is equivalent to y x 2 6 x 2 ? (1) y x 3 7
(3) y x 3 2
y x 2 6 x 3 3 2
(2) y x 3 2
(4) y x 6 10
y x 3 7
2
2
2
2
2
2
y x2 6 x 9 9 2
(1)
2
Exercise #6: Consider the quadratic function f x x 2 2 x 8 . (a) Write the function in its vertex form and identify the coordinates of the parabola’s turning point. f x x 2 2 x 1 1 8 2
f x x2 2x 1 1 8 f x x 1 9 2
Turning Point at 1, 9
2
(b) Find its x-intercepts algebraically by setting the equation you found in part (a) to zero.
x 1 9 0 2 x 1 9 x 1 3 2
x 1 3 or x 1 3 x 2 or x 4
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #3 – QUADRATIC FUNCTIONS AND THEIR ALGEBRA – LESSON #12 eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
Answer Key Name: ___________________________________
Date: _________________
SHIFTING PARABOLAS AND COMPLETING THE SQUARE ALGEBRA 2 WITH TRIGONOMETRY - HOMEWORK SKILLS 1. Which of the following equations would result from shifting y x 2 five units right and four units up? (1) y x 5 4
(3) y x 4 5
(2) y x 5 4
(4) y x 4 5
2
2
2
(1)
2
2. Which of the following represents the turning point of the parabola whose equation is y x 3 7 ? 2
(1) 3, 7
(3) 7, 3
y x 2 has been shifted 3 left and 7 down.
(2) 3, 7
(4) 3, 7
Thus, the new turning point is at 3, 7.
(4)
3. Which of the following quadratic functions would have a turning point at 6, 2 ? (1) y x 6 2
(3) y x 6 2
(2) y x 2 6
(4) y x 2 6
2
2
y x 2 would need to be shifted 6 right and 2 unit downward. Thus y x 6 2. 2
2
2
(3)
4. Which of the following is equivalent to y x 2 12 x 4 ? (1) y x 6 40
(3) y x 6 4
y x 2 12 x 6 6 4
(2) y x 6 28
(4) y x 12 148
y x 6 40
2
2
2
2
2
2
y x 2 12 x 36 36 4
(1)
2
5. In vertex form, the parabola y x 2 10 x 8 would be written as (1) y x 5 33
(3) y x 10 92
(2) y x 5 17
(4) y x 10 108
2
2
2
2
y x 2 10 x 5 5 8 2
2
y x 2 10 x 25 25 8
(2)
y x 5 17 2
6. The turning point of the parabola y x 2 5x 2 is (1) 2.5, 12.75 (2) 5, 10.5
(3) 2.5, 8.25 (4) 2.5, 17.5
y x 2 5 x 2.5 2.5 2 2
2
y x 2 5 x 6.25 6.25 2 y x 2.5 8.25 2
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #3 – QUADRATIC FUNCTIONS AND THEIR ALGEBRA – LESSON #12 eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
(3)
7. Write each of the following quadratic functions in its vertex form by completing the square. (a) y x 2 12 x 50
(b) y x 2 10 x 7
y x 2 12 x 6 6 50
y x 2 10 x 5 5 7
y x 2 12 x 36 36 50
y x 2 10 x 25 25 7
y x 6 14
y x 5 18
2
2
2
2
2
2
8. Consider the quadratic function whose equation is y x 2 6 x 40 . (a) Determine the y-intercept of this function algebraically.
(b) Write the function in its vertex form. State the coordinates of its turning point. y x 2 6 x 3 3 40 2
y 0 6 0 40 2
2
y x 2 6 x 9 9 40 y x 3 49
y 40
2
Turning Point at 3, 49
(c) Algebraically find the x-intercepts of the function by setting the equation that you found in part (b) equal to zero.
x 3 49 0 2 x 3 49
(d) Sketch a graph of the parabola, showing all relevant features found in parts (a) through (c). y
2
x
10, 0
x 3 7 x 3 7 or x 3 7 x 10 or x 4
4, 0 0, 40 3, 49
y 9. The quadratic function shown graphed to the right has the form y x 2 bx c . Determine its equation first in vertex form and then determine the values of b and c. This graph represents y x 2 after a shift of 1 unit left and 4 units down. Thus:
Vertex Form: y x 1 4 2
And: y x 1 x 1 4 y x2 x x 1 4 y x 2 2 x 3 b 2 and c 3
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #3 – QUADRATIC FUNCTIONS AND THEIR ALGEBRA – LESSON #12 eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
x
