CHAPTER 6 – TROGONOMETRY€¦ · Web view–2 4 –1 1 0 0 1 1 2 4 x. FEATURES OF THE QUADRATIC RELATION y = x2. Direction. of. Opening Up Vertex Lowest point on a parabola. The

MFM I

GRAPHING THE QUADRATIC RELATION y = x 2

The base equation that defines a quadratic functions is:

Create a table of values and plot the coordinates for y = x2.

y

x

FEATURES OF THE QUADRATIC RELATION y = x 2

Directionof Up

x y–2 4–1 10 01 12 4

UNIT 8 – Represent Quadratic Relations

y = x2

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Opening

VertexLowest point on a parabola. The

coordinate where the graph intersects (splits) the parabola in half.

(0, 0)

Maximum or Minimum

Value

The y-coordinate of the lowest point (vertex) on a parabola that opens up.

Minimum Value is 0.(min. = 0)

Axis of Symmetry

A vertical line that passes through the vertex.

(The line of symmetry.)x = 0

(y –axis)

x-interceptsThe coordinate (point) where the

parabola crosses the x-axis. (0, 0)

y

Minimum Value0 when x = 0 x-intercept

x

Vertex Axis of Symmetry (0, 0) x = 0

EXPLORING y = ax 2 1. Complete a table of values for each:

a. b. c.y = x2 y = 2x2 y = 0.5x2

–2 –2 –2–1 –1 –1


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0 0 01 1 12 2 2

2. Graph each of the functions.

a. y = x2 b. y = 2x2 c. y = 0.5x2

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6

4

2

0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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2

0

-2

-4

-6

-8

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6

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2

0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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0

-2

-4

-6

-8

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6

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2

0

-2

-4

-6

-8

3. Describe the shape of the graphs:

The graph of a quadratic equation is a PARABOLA . (U-shaped graph)

4. Compare the three parabolas. What is different about them? How are they similar?

The graphs have the same vertex. They all have different widths. (y = 2x2 is narrower than the others)

GRAPHING y = ax 2

In General:


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Examples: y = 2x2 y = x2

y = ½x2

y


When a > 0 (positive), the parabola opens upward.

When a < 0 (negative), the parabola opens downward.

AND

If a > 1 OR a < –1 there is a vertical stretch in the y-direction (the parabola narrows).

If –1 < a < 1 (a fraction or decimal) there is a vertical compression in the y-direction (the parabola widens).

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x

y = 2x2 y = ½x2

Direction of Opening

Up Up

Compression/Stretch

Stretch Compression

Vertex (0, 0) (0, 0)Axis of Symmetry x = 0 (y-axis) x = 0 (y-axis)

y = – 2x2 y =

–x 2 y = –½ x2

y

x

y = –2x2 y = –½x2

Direction of Opening Down DownCompression/Stretch Stretch Compression

Vertex (0, 0) (0, 0)Axis of Symmetry x = 0 (y-axis) x = 0 (y-axis)

THE GRAPHS OF y = x 2 AND y = x 2 + k

In General:


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The graph of y = x2 + 2 is the graph of y = x2 shifted up two units.

The graph of y = x2 – 3 is the graph of y = x2 shifted down three units.

y = x2 + 2 y = x2 y = x2 – 3

y

x

THE GRAPHS OF y = x 2 AND y = (x – h) 2

In General:


If h > 0, the graph of y = (x – h)2 is the graph of y = x2 shifted to the RIGHT by h units.

If h < 0, the graph of y = (x + h)2 is the graph of y = x2 shifted to the LEFT by h units.

THINK OPPOSITE

If the number inside the brackets is positive, move to the left. If the number inside the brackets is negative, move to the right.

When k > 0, shift upwards (y = x2 + k)

When k < 0, shift downwards (y = x2 – k)

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The graph of y = (x – 2)2 is the graph of y = x2 shifted two units to the right.

The graph of y = (x + 3)2 is the graph of y = x2 shifted three units to the left.

Examples:

y = (x + 3) 2 y = x2 y = (x – 2)2

y

x

GRAPHING y = a(x – h) 2 + k

In previous sections we investigated the effect on the graph ofy = x2 of the constants a, h, and k in the equations:


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Reflection in x-axis Horizontal Shift Left/Right)

y = a (x – h)2 + k

Vertical Stretch or Compression Vertical Shift Up/Down

EXAMPLES: Graph and state the properties for each quadratic relation

1. y = 2(x + 4)2 – 1.

2. y = ½(x + 1)2 – 2.

3. y = –2(x – 3)2 + 2.

4. y = –½x2 – 3.

Graph And State The Properties For Each Parabola

** Note: Draw each parabola separately on a grid with the original (y = x2). Complete the chart below for each function (properties).


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1) y = (x – 2)2 + 3 6) y = x2 + 1

2) y = ¼(x + 1)2 – 3 7) y = –(x + 5)2 – 3

3) y = –3x2 – 2 8) y = 2x2 + 3

4) y = –½(x + 2)2 – 1 9) y = (x – 3)2 + 2

5) y = 0.6(x – 4)2 10) y = 5(x + 4)2 – 2

Direction of

Opening

Vertex

Maximum/Minimum

Value

Equation of Axis of

Symmetry

Stretch/Compression/

Reflection

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.GRAPHING & STATING PROPERTIES - Worksheet

1) y = 2x2 – 2 2) y = (x + 1)2 3) y = x2 – 3

Direction of


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Opening:

Vertex:

Maximum/Minimum Value:

Equation of Axis of Symmetry


Reflection

4) y = –x2 + 3 5) y = (x – 1)2 + 2 6) y = –(x – 2)2 + 2

Direction of Opening:

Vertex:

Maximum/Minimum Value:

Equation of Axis of Symmetry


Reflection

GRAPHING QUADRATIC RELATIONS – SUMMARY


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y = x2 Graph is a parabola.

y = ax2

Stretches in the y-direction if a >1OR a < –1.

Shrinks (compresses) in the y-direction if –1 < a < 1.

Reflects in the x-axis if a < 0.

y = a(x – h)2

y = a(x + h)2

Shifts h units to the right. (h > 0)

Shifts h units to the left. (h < 0)

y = a(x – h)2 + k

y = a(x – h)2 – k

Shifts k units upward. (k > 0)

Shifts k units downward. (k < 0)

RATES OF CHANGE IN QUADRATIC RELATIONS

Each type of relation (linear and quadratic) has its own shape of graph.

We can examine the graph, a table of values, and equation to determine the type of relation it represents.


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2 4

2

4

6

8

10

x

y

I

For any linear relation:

The graph is a straight line. The rate of growth or decay is constant. (slope) The equation has the form: y = mx + b (Eg. y = 2x – 1) The 1st differences between consecutive y-values are constant

(slope).

x y 1st Differences0 11 3

2 5

3 7


LINEAR RELATIONS

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2

4

6

8

10

x

y

I

The 1st differences are constant.The rate of growth is constant. (Slope = 2)

For any quadratic function:

The graph is a curve called a parabola.

The rate of growth or decay is not constant, but increases or decreases by a constant amount.

The equation has the form: y = a(x – h)2 + k (Eg. y = (x + 2)2 – 3)

The 2nd differences in are constant (equal).

The 2nd differences are constant (equal).

x y 1st

Differences2nd

Differences0 11 22 53 10


QUADRATIC RELATIONS

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DIFFERENT FORMS OF THE QUADRATIC RELATION

WRITING QUADRATIC RELATIONS IN STANDARD FORM

Example #1: a) Expand and simplify y = (x – 4)2 to write the quadratic relation in standard form.

y = (x – 4)2

= (x – 4)(x – 4)

= x2 – 8x + 16

b) Determine the y-intercept.


Vertex Form: y = a(x – h)2 + k

Standard Form: y = x2 + bx + c

the constant, c, represents the y-intercept

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To find the y-intercept, substitute x = 0 and solve for y.

y = (0)2 – 8(0) + 16

= 16 (The graph will cross the y-axis at 16.)

Example #2: Rewrite y = (x + 5)2 – 4 in standard form.

y = (x + 5)(x + 5) – 4

= x2 + 5x + 5x + 25 – 4

= x2 + 10x + 25 – 4

= x2 + 10x + 21

Example #3: Determine the y-intercept for the quadratic relation y = x2 – 4x + 6.

y = (0)2 – 4(0) + 6

= 0 + 0 + 6

= 6

The graph will cross the y-axis at 6.


MFM I

DIFFERENT FORMS OF THE QUADRATIC RELATION

Example #1: Expand and simplify y = 2(x – 2)2 to write the quadratic relation in standard form.

y = 2(x – 2)2

= 2(x – 2)(x – 2)

= 2(x2 – 2x – 2x + 4)

= 2x2 – 4x – 4x + 8

= 2x2 – 8x + 8

Example #2: Rewrite y = –2(x + 3)2 – 4 in standard form AND find the y-intercept.

y = –2(x + 3)(x + 3) – 4

= –2(x2 + 3x + 3x + 9) – 4

= –2x2 – 6x – 6x – 18 – 4

= –2x2 – 12x – 22


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y-intercept Substitute x = 0 and solve for y.

y = –2(0)2 – 12(0) – 22

= –22

Example #3: Rewrite y = –½(x – 6)2 + 2 in standard form AND find the y-intercept.

y = –½(x – 6)2 + 2

= –½(x – 6)(x – 6) + 2

= –½(x2 – 6x – 6x + 36) + 2

= –½x2 + 3x + 3x – 18 + 2

= –½x2 + 6x – 16

y-intercept Substitute x = 0 and solve for y.


MFM I

y = –½(0)2 + 6(0) – 16

= 16

THE x-INTERCEPTS OF A QUADRATIC RELATION

The x-intercepts are:

The x-coordinates of the points where the graph crosses the x-axis.

Also called the ZEROS.

Example #1: Determine the zeros for y = x2 + 5x + 6.


To find the x-intercepts of a quadratic relation y = x2 + bx + c:

Factor the relation.

The result will produce the quadratic relation in the formy = (x – r)(x – s)

The x-intercepts are at x = r and x = s. (opposite of factors)

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y = x2 + 5x + 6

Factor: = (x + 2) (x + 3)

Zeros (x-intercepts) x = –2 and x = –3 (opposite 2 and 3)

From the original relation, the y-intercept is 6. To sketch the graph, plot the zeros and y-intercept and draw a smooth

parabola.Example #2: A quadratic function is given by y = x2 – x – 6.

Determine the x and y intercepts.

x-intercepts y-intercepts

y = x2 – x – 6 y = (0)2 – (0) – 6

y = (x – 3) (x + 2) = –6

x = 3 OR x = –2 The y-intercept is

The x-intercepts are: (0, –6) (3, 0) and (–2, 0)

Example #3: For the quadratic function y = x2 – 4x – 5, determine the x and y intercepts.

y-intercept:

Substitute x = 0 in the equation and solve for y.


MFM I

y = (0)2 – 4(0) – 5 The y-intercept is (0, –5).

x-intercepts:

Factor the trinomial. Then write opposite integers of factors.

y = x2 – 4x – 5

y = (x – 5)(x + 1) The x-intercepts are: (5, 0) and (–1, 0)

FORMS OF QUADRATIC RELATIONS

QuadraticForm

QuadraticModel

InformationObtained

Standard y = ax2 + bx + c Direction of Opening Stretch or Compression y-intercept

Vertex y = a(x – h)2 + k

Direction of Opening Stretch or Compression Vertex Maximum or Minimum

Value

Intercept y = a(x – r)(x – s) Zeros (x-intercepts) Axis of Symmetry

Example #1: Express y = 3x2 + 39x + 120 in intercept form.

y = 3 (x2 + 13x + 40)


MFM I

= 3 (x + 8) (x + 5)

Example #2: Find the zeros for y = –2x2 + 18.

y = –2 (x2 – 9)

= –2 (x + 3) (x – 3)

Zeros x = –3 and x = 3

P RACTICE :

A quadratic function is given by y = (x – 1)2 – 16.

a) Give the coordinates of the vertex.b) Rewrite the function in the form y = ax2 + bx + c.c) What is the y-intercept?d) Determine the x-intercepts.e) Sketch the function.

Solutions:

a) From the Standard Form: Vertex = (h, k) (1, –16)

b) Expand and simplify the standard form:

y = (x – 1)2 – 16

= (x – 1) (x – 1) – 16


MFM I

= x2 – x – x + 1 – 16

= x2 – 2x – 15

c) Use the Standard Form to determine the y-intercept: (Substitute x = 0 in the equation and solve for y)

y = (0)2 – 2(0) – 15

= –15

The y-intercept is (0, –15)

d) Use the Standard Form to determine the x-intercept: (Factor the equation then solve for x)

0 = x2 – 2x – 15

0 = (x – 5)(x + 3)

x – 5 = 0 OR x + 3 = 0

x = 5 x = –3

The x-intercepts are: (5, 0) and (–3, 0)

e) Sketch the function.


MFM I

x

y

SOLVE PROBLEMS INVOLVING QUADRATIC RELATIONS

The graph of a quadratic relation is symmetrical.

The maximum or minimum lies halfway between the zeros (x-intercepts), on the AXIS OF SYMMETRY.


Axis of Symmetry A vertical line through the vertex of a parabola.

The x-intercept of the axis of symmetry is midway between the zeros.

Must be written as an equation. eg. (x = 3 )

MFM I

Example: Find the equation of the axis of symmetry for the quadratic relation y = –3(x + 5)(x – 3).

Zeros x = –5 and x = 3

Axis of Symmetry Halfway between the zeros.

x = –1


Documents

CHAPTER 6 – TROGONOMETRY€¦ · Web view–2 4 –1 1 0 0 1 1 2 4 x. FEATURES OF THE QUADRATIC RELATION y = x2. Direction. of. Opening Up Vertex Lowest point on a parabola. The