Unit 2 Review (MPM2DG)

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Graphing Relations, Absolute Value, Transformations Linear Relationships Completing the equation using the method o The equation is of the form = β‹― o Since the point (___,___) is on the line, it must satisfy its equation. o (Simplify and isolate missing variable) o Answer the original question: ∴ the equation (of the line) is = β‹― Inequalities Greater-than points are always above the line Less-than points are always below the line If the inequality is an β€œor equal to” (≀β‰₯): o Solid line If not equal to: o Dotted line Intersection and union o Intersection = overlap No solid lines inside shaded region allowed o Union = everything No lines inside shaded region Absolute Value Size of number (distance from 0 on a number line) o |3| = |βˆ’3| = 3 o Brackets Whenever possible, evaluate contents first Result of ||=2 is a set or an OR statement o {| = 2 OR = βˆ’2, βˆˆβ„} o = 2 OR = βˆ’2 || cannot be negative o ||= βˆ’3 has no solution To solve ||= , β‰₯ 0 o = βˆ’ OR = To solve |βˆ’ 9| = 9 o βˆ’ 9= βˆ’9 OR βˆ’ 9=9 o = 0 OR = 18 always reorder so that the greater is on the right, as on a number line

description

A review of graphs of relations, absolute value, and transformations.

Transcript of Unit 2 Review (MPM2DG)

Graphing Relations, Absolute Value, Transformations

Linear Relationships Completing the equation using the method

o The equation is of the form 𝑦𝑦 = β‹― o Since the point (___,___) is on the line, it must satisfy its equation. o (Simplify and isolate missing variable) o Answer the original question: ∴ the equation (of the line) is 𝑦𝑦 = β‹―

Inequalities Greater-than points are always above the line Less-than points are always below the line If the inequality is an β€œor equal to” (≀β‰₯):

o Solid line If not equal to:

o Dotted line Intersection and union

o Intersection = overlap No solid lines inside shaded region allowed

o Union = everything No lines inside shaded region

Absolute Value Size of number (distance from 0 on a number line)

o |3| = |βˆ’3| = 3 o Brackets

Whenever possible, evaluate contents first Result of |π‘₯π‘₯| = 2 is a set or an OR statement

o {π‘₯π‘₯|π‘₯π‘₯ = 2 OR π‘₯π‘₯ = βˆ’2,π‘₯π‘₯ ∈ ℝ} o π‘₯π‘₯ = 2 OR π‘₯π‘₯ = βˆ’2

|π‘₯π‘₯| cannot be negative o |π‘₯π‘₯| = βˆ’3 has no solution

To solve |π‘₯π‘₯| = π‘Žπ‘Ž,π‘Žπ‘Ž β‰₯ 0 o π‘₯π‘₯ = βˆ’π‘Žπ‘Ž OR π‘₯π‘₯ = π‘Žπ‘Ž

To solve |π‘₯π‘₯ βˆ’ 9| = 9 o π‘₯π‘₯ βˆ’ 9 = βˆ’9 OR π‘₯π‘₯ βˆ’ 9 = 9 o π‘₯π‘₯ = 0 OR π‘₯π‘₯ = 18 always reorder so that the greater is on the right, as on a number line

Absolute value inequalities

For |π‘₯π‘₯| > π‘Žπ‘Ž,π‘Žπ‘Ž β‰₯ 0,π‘Žπ‘Ž ∈ ℝ π‘₯π‘₯ < βˆ’π‘Žπ‘Ž OR π‘₯π‘₯ > π‘Žπ‘Ž

For |π‘₯π‘₯| < π‘Žπ‘Ž,π‘Žπ‘Ž β‰₯ 0,π‘Žπ‘Ž ∈ ℝ (double less-than) βˆ’π‘Žπ‘Ž < π‘₯π‘₯ < π‘Žπ‘Ž

|π‘₯π‘₯ + 𝑦𝑦| = | βˆ’ π‘₯π‘₯ βˆ’ 𝑦𝑦|

Absolute Value Graphs Square diamonds produced with |π‘₯π‘₯| + |𝑦𝑦| = π‘šπ‘š with m being the radius from the centre to any vertex.

Centre can be transformed inside absolute value.

Double-V opening to the left and right formed with

|π‘₯π‘₯| βˆ’ |𝑦𝑦| = π‘šπ‘š,π‘šπ‘š > 0 touching the points (Β±π‘šπ‘š, 0) and 2π‘šπ‘š apart.

Symmetrical; line of symmetry may be transformed.

Double-V opening up and down formed with |π‘₯π‘₯| βˆ’ |𝑦𝑦| = π‘šπ‘š,π‘šπ‘š < 0

with vertices at (0, Β±π‘šπ‘š).

𝑦𝑦 = π‘šπ‘š|π‘₯π‘₯|,π‘šπ‘š > 0 produces a single-V opening up, where the slopes of the rays are Β±π‘šπ‘š.

𝑦𝑦 = π‘šπ‘š|π‘₯π‘₯|,π‘šπ‘š < 0 produces a single-V opening down, with slopes Β±π‘šπ‘š.

Backwards rule o 𝑦𝑦 = |π‘₯π‘₯ + 3| moves the relation to the left by 3 units. 𝑦𝑦 = |π‘₯π‘₯ βˆ’ 3| moves it to the right. o 𝑦𝑦 + 3 = |π‘₯π‘₯| moves the relation down by 3 units. 𝑦𝑦 βˆ’ 3 = |π‘₯π‘₯| moves it up.

|π‘₯π‘₯ βˆ’π‘šπ‘š| βˆ’ |𝑦𝑦 βˆ’ 𝑛𝑛| = 1 produces a double-V with the vertical line of symmetry moved π‘šπ‘š to the right, and the horizontal line of symmetry moved up by 𝑛𝑛.

The distance away from the vertical line of symmetry is 1, and any increase is reflected by an equal movement of the Vs away from the line of symmetry.

|π‘₯π‘₯|π‘šπ‘š

+|𝑦𝑦|𝑛𝑛

= 1

produces a diamond with horizontal β€˜radius’ π‘šπ‘š and vertical β€˜radius’ 𝑛𝑛. The centre may be transformed by means of the backwards rule.

|π‘₯π‘₯|π‘šπ‘šβˆ’

|𝑦𝑦|𝑛𝑛

= 1

produces a set of double-Vs with slopes Β± π‘›π‘›π‘šπ‘š

and π‘šπ‘š away from the

vertical line of symmetry.

|π‘₯π‘₯|π‘šπ‘šβˆ’

|𝑦𝑦|𝑛𝑛

= βˆ’1

produces a set of double-Vs opening up and down, 𝑛𝑛 away from the horizontal line of symmetry. This equation, combined with the equation above, form shapes that indicate the diamond formed with the same variables.

Simple Transformations Mappings

o Written (π‘₯π‘₯, 𝑦𝑦) β†’ (__, __) o or with point names 𝑃𝑃(π‘₯π‘₯,𝑦𝑦) β†’ 𝑃𝑃′(__, __) o May reference initial values (e.g. (π‘₯π‘₯,𝑦𝑦) β†’ (2π‘₯π‘₯, 2𝑦𝑦))

Translations o Of the form

𝑃𝑃(π‘₯π‘₯,𝑦𝑦) β†’ 𝑃𝑃′(π‘₯π‘₯ + π‘Žπ‘Ž,𝑦𝑦 + 𝑏𝑏) where π‘Žπ‘Ž, 𝑏𝑏 ∈ ℝ o Remain the same: shape, orientation, size, slopes, sense o Change: location

Reflections o When drawing, always indicate mirror line, perpendicular construction lines o Remain the same: size, shape o Change: orientation, (location), sense, slopes of sides

o Reflections in the x-axis 𝑃𝑃(π‘₯π‘₯,𝑦𝑦) β†’ 𝑃𝑃′(π‘₯π‘₯,βˆ’π‘¦π‘¦)

o Reflections in the y-axis 𝑃𝑃(π‘₯π‘₯,𝑦𝑦) β†’ 𝑃𝑃′(βˆ’π‘₯π‘₯,𝑦𝑦)

o Reflections in 𝑦𝑦 = βˆ’π‘₯π‘₯ 𝑃𝑃(π‘₯π‘₯,𝑦𝑦) β†’ 𝑃𝑃′(βˆ’π‘¦π‘¦,βˆ’π‘₯π‘₯)

o Reflections in 𝑦𝑦 = π‘₯π‘₯ 𝑃𝑃(π‘₯π‘₯,𝑦𝑦) β†’ 𝑃𝑃′(𝑦𝑦, π‘₯π‘₯)

Rotations o Positive angles always clockwise o Negative angles always counter-clockwise o 90Β° rotation

𝑃𝑃(π‘₯π‘₯,𝑦𝑦) β†’ 𝑃𝑃′(βˆ’π‘¦π‘¦, π‘₯π‘₯) Apply pattern for 180Β° and 270Β° rotations

o Remain the same: shape, sense, size o Change: orientation, slopes, location

Dilatations o k value is the multiplier for the distances from the centre to any point o Negative k values: measure in other direction o Remain the same: slopes, orientation (+ k), sense, shape o Change: location, area/size, orientation (- k)

Composition of Transformations o E.g. 𝑅𝑅π‘₯π‘₯ ∘ 𝑇𝑇0,2 o The last transformation is done first

The Method for transformations

Let (π‘₯π‘₯,𝑦𝑦) β†’ (π‘₯π‘₯ βˆ’ 2,𝑦𝑦 + 2) = (π‘₯π‘₯β€² ,𝑦𝑦′)

∴ οΏ½π‘₯π‘₯β€² = π‘₯π‘₯ βˆ’ 2𝑦𝑦′ = 𝑦𝑦 + 2 β‡’ οΏ½π‘₯π‘₯ = π‘₯π‘₯β€² + 2

𝑦𝑦 = 𝑦𝑦′ βˆ’ 2οΏ½οΏ½

Since (π‘₯π‘₯,𝑦𝑦) is on the original line, it must satisfy its equation. (substitute second set of equations into original equation of line)

𝑦𝑦′ βˆ’ 2 =32

(π‘₯π‘₯β€² + 2) + 3

𝑦𝑦′ βˆ’ 2 =32π‘₯π‘₯β€² + 3 + 3

𝑦𝑦′ =32π‘₯π‘₯β€² + 8

(now drop primes for the image equation)

∴ The image equation is 𝑦𝑦 =32π‘₯π‘₯ + 8.

Other Transformations One-way stretches

o Multiplication of x value only or y value only o E.g. (π‘₯π‘₯,𝑦𝑦) β†’ (2π‘₯π‘₯,𝑦𝑦)

2-way stretches o Multiplication of x and y (if same factor, then it is a dilatation) o E.g. (π‘₯π‘₯,𝑦𝑦) β†’ (3π‘₯π‘₯, 4𝑦𝑦)

Shears o Adding x to y or y to x o Horizontal shear: (π‘₯π‘₯,𝑦𝑦) β†’ (π‘₯π‘₯ + 𝑦𝑦,𝑦𝑦) or even (π‘₯π‘₯ + 2𝑦𝑦, 𝑦𝑦) o Vertical shear: (π‘₯π‘₯,𝑦𝑦) β†’ (π‘₯π‘₯, π‘₯π‘₯ + 𝑦𝑦)

Non-Standard Transformations (not at the origin)

Any rotation or dilatation about a point not at the origin, or a reflection on a non-standard line of symmetry may be solved by translating the problem to the origin or standard locations, applying the transformation, and reversing the translation.

e.g.

rotation of 90Β° about (3,5) π‘‡π‘‡βˆ’3,βˆ’5: (π‘₯π‘₯,𝑦𝑦) β†’ (π‘₯π‘₯ βˆ’ 3,𝑦𝑦 βˆ’ 5) 𝑅𝑅90Β°: (π‘₯π‘₯ βˆ’ 3,𝑦𝑦 βˆ’ 5) β†’ (βˆ’π‘¦π‘¦ + 5, π‘₯π‘₯ βˆ’ 3) 𝑇𝑇3,5: (βˆ’π‘¦π‘¦ + 5, π‘₯π‘₯ βˆ’ 3) β†’ (βˆ’π‘¦π‘¦ + 8, π‘₯π‘₯ + 2) ∴ The rotation of point (π‘₯π‘₯,𝑦𝑦) about (3,5) is (π‘₯π‘₯,𝑦𝑦) β†’ (βˆ’π‘¦π‘¦ + 8, π‘₯π‘₯ + 2).