What Does Your Quadratic Look Like? EXAMPLES

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What Does Your Quadratic Look Like?

EXAMPLES

1. An equation such as “y = x2 – 4x + 1” describes a type of “function” known as a

“quadratic function.” Review with students that a “function” is a “relation” in which each element of the “domain” is paired with exactly one element of the “range.” Remind them of the “vertical line test” for “functions”.

2.

3. Graphs of “quadratic functions” have certain common characteristics. For instance, they all have a general shape called a “parabola.” Note: You may wish to demonstrate using a three-dimensional model of a double cone. Explain that a parabola is a conic section formed by the intersection of a double cone and a plane that is parallel to the surface of the cone.

4. The table and graph below illustrate some other common characteristics of “quadratic functions.” x x2 – 4x + 1 y -1 (-1)2 – 4(-1) + 1 6 0 02 – 4(0) + 1 1 1 12 – 4(1) + 1 -2 2 22 – 4(2) + 1 -3 3 32 – 4(3) + 1 -2 4 42 – 4(4) + 1 1 5 52 – 4(5) + 1 6

Definition of a Quadratic Function

A “quadratic function” is a function that can be described by an equation of the form “y = ax2 + bx + c,” where a ≠ 0.

Notice the matching values in the y-column.

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5. Notice that in the y-column of the table, “-3” does not have a matching value. Also, “-3” is the “y-coordinate” of the lowest point on the graph of “y = x2 – 4x + 1.” For the graph of “y = x2 – 4x + 1,” The lowest point, called the

“minimum point,” has the coordinates “(2, -3).” The “minimum” or “maximum” point of a “parabola” is often called the “vertex.”

6. The vertical line containing the “minimum point” (2, -3), is called the “axis of

symmetry.” The equation of the “axis of symmetry” for the graph above is “x = 2.”

7. If the graph of any “quadratic” function is folded along the ”axis of symmetry,” the two halves coincide. In other words, the two halves of the “parabola” are “symmetric.” 8. Example: GraphÎy = –x2 + 2x + 3 x -x2 + 2x + 3 y -2 -(-2)2 + 2(-2) + 3 -5 -1 -(-1)2 + 2(-2) + 3 0 0 -(0)2 + 2(-2) + 3 3 1 -(1)2 + 2(-2) + 3 4 2 -(2)2 + 2(-2) + 3 3 3 -(3)2 + 2(-2) + 3 0 4 -(4)2 + 2(-2) + 3 -5 X y = x2 – 4x + 1 (2, -3) Axis of symmetry x = 2 Axis of symmetry (1, 4)

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9. The graph of “y = -x2 + 2x + 3” opens downward. The equation of the “axis of symmetry” is “x = 1.” The graph has a highest point, or “maximum point,” at

(1, 4).

10. In general, a parabola will open upward and have a “minimum” point when the coefficients of “y” and “x2” have the same sign. It will open downward and have a “maximum” point when the coefficients of “y” and “x2” have the opposite sign. The “maximum” or “minimum” point of the graph always lies on the “axis of

symmetry.”

11. Notice in the graph of “y = -x2 + 2x + 3,” the “axis of symmetry” is halfway

between any two points have the same “y-coordinate.” Consider the points on the graph whose coordinates are (-1, 0) and (3, 0). From these coordinates, the

equation of the “axis of symmetry” may be found as shown below: 12. In general, the equation of the “axis of symmetry for the graph of a

“quadratic” function can be found by using the following rule

13. Example: Find the equation of the axis of symmetry and the

coordinates of the maximum or minimum point of the graph of y =x2 – x – 6. Then use the information to draw the graph. First, find the equation of the axis of symmetry.

The equation of the axis of symmetry is x = 2 1

.

Next, find the coordinates of the maximum or minimum point. Since the coefficients of “y” and “x2” have the same sign, the graph of the function has a minimum point. The minimum point lies on the axis of symmetry. Since the axis of symmetry is x =

2 1 , the x = 2 3 1+ − x = 1 Equation of Axis of Symmetry

The equation of the axis of symmetry for y = ax2 + bx + c, where a ≠0, is x = a b 2 − . x = a b 2 − x = 1 2 1 • − − x = 2 1 Remind students that “a” is the coefficient of “x2” and “b” is the coefficient of “x.”

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minimum point will have an x-coordinate of 2 1

. Find the y-coordinate by substituting

2 1

for “x” in y = x2 – x – 6.

The coordinates of the minimum point are ) 4 25 , 2 1 ( − .

Then, construct a table. For the values of “x” choose some integers greater than 2 1

, and some less than

2 1

. This insures that points on both sides of the axis of symmetry are plotted. Use this information to draw the graph.

x x2 – x - 6 y -2 (-2)2 – (-2) - 6 0 -1 (-1)2 – (-2) – 6 -4 0 (0)2 – (-2) – 6 -6 1 (1)2 – (-2) – 6 -6 2 (2)2 – (-2) – 6 -4 3 (3)2 – (-2) – 6 0 y = x2 – x – 6 6 2 1 ) 2 1 ( 2− − = y 6 2 1 4 1 − − = y 4 25 − = y Minimum point ) 4 25 , 2 1 ( − Axis of symmetry x = 2 1

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14. Example: Find the coordinates of the maximum point for the graph of y = -2x2 – 8x + 9.

Since the coefficients of “y” and “x2” have different signs, the graph of the function as a maximum point.

x = a b 2 − x = ) 2 ( 2 8 − − − x = –2

First, find the equation of the axis of symmetry. Use a = –2 and b = –8.

The equation of the axis of symmetry is x = –2

Since the maximum point lies on the axis of symmetry, substitute “–2” for “x” in y = –2x2 – 8x + 9.

y = -2(-2)2 – 8(-2) + 9 y = -8 + 16 + 9 y = 17

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WHAT DOES YOUR QUADRATIC LOOK LIKE?

WORKSHEET

Find the equation of the axis of symmetry and the coordinates of the maximum or

minimum point of the graph of each quadratic function. Students will write out a table for each problem and then draw the graph.

1. y = -x2 + 5x + 6 2. y = x2 + 2x 3. y = 3x2 – 6x + 5 4. y = x2 – 4x – 5 5. y = x2 – x – 6 Name:____________________ Date:_________ Class:____________________

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WHAT DOES YOUR QUADRATIC LOOK LIKE?

WORKSHEET KEY

Find the equation of the axis of symmetry and the coordinates of the maximum or minimum point of the graph of each quadratic function. Then draw the graph.

1. y = -x2 + 5x + 6

First, find the equation of the axis of symmetry.

The equation of the axis of symmetry is x = 2 5

.

Next, find the coordinates of the maximum or minimum point. Since the coefficients of “y” and “x2” have different signs, the graph of the function has a maximum point. The maximum point lies on the axis of symmetry. Since the axis of symmetry is x =

2 5

, the maximum point will have an x-coordinate of

2 5

. Find the y-coordinate by substituting

2 5

for “x” in y = -x2 + 5x + 6.

The coordinates of the minimum point are ) 4 49 , 2 5 ( .

, and some less than

2 5

x = a b 2 − x = 1 2 5 − • − x = 2 5 Remind students that “a” is the coefficient of “x2” and “b” is the coefficient of “x.” y = -x2 + 5x + 6 6 ) 2 5 ( 5 ) 2 5 ( 2 + + − = y 6 2 25 4 25 + + − = y 4 49 = y

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x -x2 + 5x + 6 y -2 -(-2)2 + 5(-2) + 6 -8 -1 -(-1)2 + 5(-1) + 6 0 0 -(0)2 + 5(0) + 6 6 1 -(1)2 + 5(1) + 6 10 2 -(2)2 + 5(2) + 6 12 3 -(3)2 + 5(3) + 6 12 4 -(4)2 + 5(4) + 6 10 5 -(5)2 + 5(5) + 6 6 6 -(6)2 + 5(6) + 6 0 2. y = x2 + 2x

First, find the equation of the axis of symmetry. The equation of the axis of symmetry is x = -1.

Next, find the coordinates of the maximum or minimum point. Since the coefficients of “y” and “x2” have the same sign, the graph of the function has a minimum point. The minimum point lies on the axis of symmetry. Since the axis of symmetry is x = -1, the minimum point will have an x-coordinate of -1. Find the y-coordinate by substituting –1 for “x” in y = x2 + 2x.

The coordinates of the minimum point are (-1, -1).

Then, construct a table. For the values of “x” choose some integers greater than -1, and some less than -1. This insures that points on both sides of the axis of symmetry are plotted. Use this information to draw the graph.

x = a b 2 − x = 1 2 2 • − x = -1 Remind students that “a” is the coefficient of “x2” and “b” is the coefficient of “x.” y = x2 + 2x ) 1 ( 2 ) 1 (− 2 + − = Y y = -1

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x x2 + 2x y -2 (-2)2 + 2(-2) 0 -1 (-1)2 + 2(-1) -1 0 (0)2 + 2(0) 0 1 (1)2 + 2(1) 3 3. y = 3x2 – 6x + 5

First, find the equation of the axis of symmetry. The equation of the axis of symmetry is x = 1.

Next, find the coordinates of the maximum or minimum point. Since the coefficients of “y” and “x2” have the same sign, the graph of the function has a minimum point. The minimum point lies on the axis of symmetry. Since the axis of symmetry is x = 1, the minimum point will have an x-coordinate of 1. Find the y-coordinate by substituting 1 for “x” in y = 3x2 – 6x + 5.

The coordinates of the minimum point are (1, 2).

Then, construct a table. For the values of “x” choose some integers greater than 1, and some less than 1. This insures that points on both sides of the axis of symmetry are plotted. Use this information to draw the graph.

x 3x2 – 6x + 5 y -1 3(-1)2 - 6(-1) + 5 14 0 3(0)2 - 6(-0) + 5 5 1 3(1)2 - 6(1) + 5 2 2 3(2)2 - 6(2) + 5 5 3 3(3)2 - 6(3) + 5 14 x = a b 2 − x = 3 2 6 • − − x = 1 Remind students that “a” is the coefficient of “x2” and “b” is the coefficient of “x.” y = 3x2 – 6x + 5 y = 3(1)2 – 6(1) + 5 y = 2

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4. y = x2 – 4x – 5

First, find the equation of the axis of symmetry. The equation of the axis of symmetry is x = 2.

Next, find the coordinates of the maximum or minimum point. Since the coefficients of “y” and “x2” have the same sign, the graph of the function has a minimum point. The minimum point lies on the axis of symmetry. Since the axis of symmetry is x = 2, the minimum point will have an x-coordinate of 2. Find the y-coordinate by substituting 2 for “x” in y = x2 – 4x – 5.

The coordinates of the minimum point are (2, –9).

Then, construct a table. For the values of “x” choose some integers greater than 2, and some less than 2. This insures that points on both sides of the axis of symmetry are plotted. Use this information to draw the graph.

x x2 – 4x – 5 y 0 (0)2 - 4(0) – 5 –5 1 (1)2 - 4(1) – 5 -8 2 (2)2 - 4(2) – 5 -9 3 (3)2 - 4(3) – 5 -8 4 (4)2 - 4(4) – 5 -5 x = a b 2 − x = 1 2 4 • − − x = 2 Remind students that “a” is the coefficient of “x2” and “b” is the coefficient of “x.” y = y = x2 – 4x – 5 y = (2)2 – 4(2) – 5 y = –9

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5. y = x2 – x – 6

First, find the equation of the axis of symmetry.

The equation of the axis of symmetry is x = 2 1

.

Next, find the coordinates of the maximum or minimum point. Since the coefficients of “y” and “x2” have the same sign, the graph of the function has a minimum point. The minimum point lies on the axis of symmetry. Since the axis of symmetry is x =

2 1

, the minimum point will have an x-coordinate of

2 1

. Find the y-coordinate by substituting 2 1 for “x” in y = x2 – x – 6.

The coordinates of the minimum point are ( 2 1 , 4 3 6 − ).

, and some less than

2 1

x x2 – x – 6 y 0 (0)2 - (0) – 6 –6 1 (1)2 - (1) – 6 -6 2 (2)2 - (2) – 6 -4 3 (3)2 - (3) – 6 0 -2 (-2)2 - (-2) – 6 0 x = a b 2 − x = 1 2 1 • − − x = 2 1 Remind students that “a” is the coefficient of “x2” and “b” is the coefficient of “x.” y = x2 – x – 6 y = ( 2 1 )2 – 4( 2 1 ) – 5 y = 4 3 6 −

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WHAT DOES YOUR QUADRATIC LOOK LIKE CHECKLIST

1. On questions, (1 – 5), did the student find the axis of symmetry?

a. All five (25 points)

b. Four of the five (20 points) c. Three of the five (15 points) d. Two of the five (10 points) e. One of the five (5 points)

2. On questions, (1 – 5), did the student find the coordinates of the maximum or minimum point of the graph?

a. All five (25 points)

3. On questions, (1 – 5), did the student write out a table for the graph? a. All five (25 points)

4. On questions, (1 – 5), did the student draw the graph correctly? a. All five (25 points)

b. Four of the five (20 points) c. Three of the five (15 points) d. Two of the five (10 points) e. One of the five (5 points) Total Number of Points _________ A 90 points and above

B 80 points and above C 70 points and above D 60 points and above F 59 points and below Student Name: __________________ Date: ______________

Any score below C needs