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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
MFM I
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
UNIT 8 – Represent Quadratic Relations
MFM I
0 0 01 1 12 2 2
2. Graph each of the functions.
a. y = x2 b. y = 2x2 c. y = 0.5x2
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
2
0
-2
-4
-6
-8
86420-2-4-6
6
4
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:
UNIT 8 – Represent Quadratic Relations
MFM I
Examples: y = 2x2 y = x2
y = ½x2
y
UNIT 8 – Represent Quadratic Relations
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).
MFM I
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:
UNIT 8 – Represent Quadratic Relations
MFM I
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:
UNIT 8 – Represent Quadratic Relations
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)
MFM I
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:
UNIT 8 – Represent Quadratic Relations
MFM I
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).
UNIT 8 – Represent Quadratic Relations
MFM I
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
UNIT 8 – Represent Quadratic Relations
MFM I
Opening:
Vertex:
Maximum/Minimum Value:
Equation of Axis of Symmetry
Stretch/Compression/
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
Stretch/Compression/
Reflection
GRAPHING QUADRATIC RELATIONS – SUMMARY
UNIT 8 – Represent Quadratic Relations
MFM I
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.
UNIT 8 – Represent Quadratic Relations
MFM
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
UNIT 8 – Represent Quadratic Relations
LINEAR RELATIONS
MFM
2 4
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
UNIT 8 – Represent Quadratic Relations
QUADRATIC RELATIONS
MFM I
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.
UNIT 8 – Represent Quadratic Relations
Vertex Form: y = a(x – h)2 + k
Standard Form: y = x2 + bx + c
the constant, c, represents the y-intercept
MFM I
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.
UNIT 8 – Represent Quadratic Relations
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
UNIT 8 – Represent Quadratic Relations
MFM I
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.
UNIT 8 – Represent Quadratic Relations
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.
UNIT 8 – Represent Quadratic Relations
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)
MFM I
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.
UNIT 8 – Represent Quadratic Relations
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)
UNIT 8 – Represent Quadratic Relations
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
UNIT 8 – Represent Quadratic Relations
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.
UNIT 8 – Represent Quadratic Relations
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.
UNIT 8 – Represent Quadratic Relations
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
UNIT 8 – Represent Quadratic Relations