Coordinate Geometry

Perpendicular lines
Two lines or linear equations are perpendicular if their slopes are negative reciprocals of each other (opposite signs and upside down):

m1 = − 1/m2

Example: 5x − y = 8 and 5x = − x + 3

Let's check if these two functions are Perpendicular:
- first turn the fucntions into a slope intercept form (regular form of linear equations, in oder words solve for y): y = mx + b with m = slope

5x − y = 8 − y = 8 − 5x y = 5x − 8 The slope of this line is m1 = 5 or m1 = 5/1

5y = − x + 3 y = − 1/5·x − 3/5 The slope of this line is m2 = − 1/5

- then compare their slopes or gradients:
As you can notice m₁ = ⁵/₁ is exactly the negative reciprocal of m² = − ¹/₅ and vice versa.

⇒ Therefore, the lines 5x − y = 8 and 5x = − x + 3 are perpendicular!

Parallel lines
Two lines or linear equations are parallel if their slopes are equal (they have the same slopes). Since they'll never intersect, they continue forever without touching (assuming that these lines are on the same plane).

m1 = m2

Example:
One line passes through the points p₁(–1, –2) and p₂(1, 2); another line passes through the points q₁(–2, 0) and q₂(0, 4).

Our answer to this question is based on the calculation of slopes of both lines (functions).

⇒ With identical slopes, these lines are parallel!

Neither Parallel nor Perpendicular lines
Two lines or linear equations are neither parallel nor perpendicular if their slopes are neither the same and nor negative reciprocals of each other.

m1 ≠ m2 and m1 ≠ − 1/m2

Example:
A line passes through the points p₁(–4, 2) and p₂(0, 3); another line passes through the points q₁(–3, –2) and q₂(3, 2). Are these lines parallel, perpendicular, or neither?

These slope values are not the same, so the lines are not parallel. The slope values are not negative reciprocals either, so the lines are not perpendicular. ⇒ Then the answer is "neither".!

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Solving equations - miscellaneous exercices

Possible Solutions

1. Solve the simultaneous equations
    x + y = 2 (1)
    x² + 2y² = 11 (2)

It's a system of equations, consisting of a linear and a quadratic equation. We will solve it by Substitution, the only method that suits here.
Important: It really doesn't matter what equation or what variable you pick, the answer will the same. "Kinda vice versa" ...cool, ain't it?

- first solve (1) for y:
⇒ y = 2 − x (you can call it (3), just to put everything in order)

- then plug (3) into (2)
⇒ x² + 2(2 − x)² = 11 (you can notice only one unknown variable left, makes it easier!)

- expand:
⇒ x² + 2(4 −4x + x²) = 11 (take your time and you'll come to the following result)
⇔ 3x² − 8x − 3 = 0  |  ÷3 (divide the whole equation by 3, on both side please!)
⇒ x² − ⁸/₃x − 1 = 0 (quadratic form)

- now you can solve it for: (by completing the square or what ever method you know!)
⇒ x² − ⁸/₃x + (⁴/₃)² − 1 − (⁴/₃)²
⇒ (x − ⁴/₃)² = 1 + ¹⁶/₉ = ²⁵/₉ (square root it)
⇒ x − ⁴/₃ = ± ⁵/₃
⇔ x₁ = − ⁵/₃ + ⁴/₃ = − ¹/₃,     x₁ = − ¹/₃
     or
     x₂ = ⁵/₃ + ⁴/₃ = 3,     x₂ = 3

- then you can replace both value of x in the equation (3) to find the values of y:
⇔ y₁ = 2 − (−¹/₃) = 2 + ¹/₃ = ⁷/₃,     y₁ = ⁷/₃
    y₂ = 2 − 3 = − 1,     y₂ = − 1

⇒ Then the solution is (x, y) = (−¹/₃ , ⁷/₃) or (3, −1).

2. x² − 10x + 17 (complete the square)
(x² − 10x + 5²) − 5² + 17
⇒ (x − 5)² − 8, with a = − 5 and b = − 8.
The least possible value that f(x) can take is − 8 (f(x) = y = −8) and its corresponding x value is − 5 (x = − 5).


3. Let's solve these two equations (system of equations) by substitution
    2x + y = 3 (1)
    2x² - xy = 10 (2)

- first solve (1) for y (kinda isolating y)
    ⇒ y = 3 − 2x (3)

- plug (3) into (2)
    2x² −x(3 − 2x) = 10 (expand it)
    2x² − 3x + 2x² = 10
    4x² − 3x − 10 = 0 (factorize 4)
    4(x² − ³/₄x − ¹⁰/₄) = 0 (complete the square)
    (x − ³/₄x + ³/₈)² − (³/₈)² − ¹⁰/₄
       (x − ³/₈)² = ^13²/_8² (square root it)
       x − ³/₈ = ± ¹³/₈

⇒
       x₁ = − ¹³/₈ + ³/₈ = − ⁵/₄,     x₁ = − ⁵/₄
         or
       x₂ = ¹³/₈ + ³/₈ = 2,    x₂ = 2

Substitute the values of x₁ and x₂ in (3) (kinda pluggin back into...) to find the two values of y:
Do you remember y = 3 − 2x (3)? ...great just askink! :-)

what gives u:
                 - for x = − ⁵/₄,          y = 3 − 2(− ⁵/₄) = ¹¹/₂
                 - for x = 2,               y = 3 − 2(2) = − 1

⇒ Then the solution is (x, y) = (−⁵/₄ , ¹¹/₂) or (2, −1).

4. 2x² − kx + 8 = 0 (divide it by 2 to bring it to a regular quadratic equation, without the factor 2 at x²)
⇒ x² − ^k/₂·x + 4 = 0 (complete the square)
          x² − ^k/₂·x + (^k/₄·x)² − (^k/₄·x)² + 4 = 0
          x² − ^k/₂·x + (^k/₄·x)² = (^k/₄·x)² - 4
Now compare this two equations:
     x² − ^k/₂x + 4 = 0
         and
     x² − ^k/₂·x + (^k/₄·x)² = (^k/₄·x)² - 4

as you can see by identifying:
x² = x² is a true statement;
it goes the same for − ^k/₂·x = − ^k/₂·x
(^k/₄)² = 4
and (^k/₄)² - 4 = 0 impossible!

Therefore you can work out: (^k/₄)² = 4  |   √ (square root it)
⇒ ^k/₄ = ± 2 and k = ± 8

★
You can also use the Discriminant (Δ):
The equation again: 2x² − kx + 8 = 0
by identification:
     Δ = b² − 4·a·c
with a = 2; b = −k and c = 8
Δ = k² − 4 · 2 · 8 = 0
⇒ k² − 64 = 0
⇒ k² = 64   |   √ (square root it)
     what gives you: k = ± 8


5. f(x) = (2x + 3)(x + 4)     (expand it)
      = 2x² − 5x − 7    ÷2 (divide it by 2)
      = x² − ⁵/₂·x − ⁷/₂       (complete the square)
      = x² − ⁵/₂·x + (⁵/₄)² − (⁵/₄)² − ⁷/₂
by working it out, you'll get:
(x − ⁵/₄)² = ⁸¹/₁₆   |   √ (square root it)
⇒ x = ⁵/₄ ± ⁹/₄ ⇒ x = ⁵/₄ − ⁹/₄ = − 1 or x = ⁵/₄ + ⁹/₄ = ⁷/₂
           x = − 1
                or
           x = ⁷/₂


6. 1. x² − (6√3)·x + 24 = 0
   x² − (6√3)·x + (3√3)² − (3√3)² + 24 = 0
   work it out
   to get: (x − 3√3)² = 3   |   √ (square root it)
   ⇒ x − 3√3 = ± √3
   
   Therefore the answer in terms of surds is:
              x = 2√3
              or
              x = 4√3
   
   
   2. x⁴ − (6√3)·x² + 24 = 0

Coming soon...

Lösung zu Potenzen

1) Berechne folgende Potenzterme

a)	(2cd)3 = 23c3d3 = 8c3d3	b)	(ab)2 = a2b2	c)	− (ab)2 = − a2b2
d)	− (−2ad)3 = − (−8a3d3) = 8a3d3	e)	− (2bc)3 = − 8b3c3	f)	− (ac)3 = − a3c3

2) Vereinfache

a) 103/10−2 = 105 = 100000	b) 2,54/2,54 = 2,50 = 1	c) 2,50/3−4 = 34 = 81
d) 50,5/(−0,5)0 = 50,5 = √5 ≈ 2,24	e) 2−3/2−9 = 26 = 64	f) 243/83 = 33 = 27

... coming soon ...

Potenzengesetze, Potenzen und Potenzeregeln 1

Aufgaben:

1

2

3

4

5

1) Berechne folgende Potenzterme

a)	(2cd)3	b)	(ab)2	c)	− (ab)2
d)	− (−2ad)3	e)	− (2bc)3	f)	− (ac)3

Lösung

2) Vereinfache

a)	103 10−2	b)	2,54 2,54	c)	2,50 3−4
d)	50.5 − 0.50	e)	2−3 2−9	f)	243 83

3) Berechne folgende Potenzterme




4) Berechne folgende Potenzterme




5) Berechne folgende Potenzterme

Lösung zur Trigonometrie (2)

Lösung 3:

Abb. Symmetrisches Trapez

Gegeben: a = 9,2cm, b = 4,0cm, α = 40°
Für ein Symmetrisches Trapez gilt:
• b = d = 4,0cm, AF = GB , FG = DC (c), FD (h) =GC
• α = β = 40°, γ = δ (δ = delta) und α + β + γ + δ = 360°
• Flächeninhalt A = ^{(a + c)·h}/₂
Gesucht: c, h, γ und A
Bekannt:
α + β + γ + δ = 360°
⇒ 40° + 40° + 2γ = 360°  |   −80°
⇒ 2γ = 280   |   ÷2;
⇒ γ = 140°     γ = 140°
γ = δ ⇒ δ = 140°

sin α = ^h/_d  |   ·d
⇒ h = d·sin 40° = 4,0·sin 40°,     h = 2,57cm

tan α = ^h/_AF  |   ·(−1)
⇒ ¹/_{tan α} = ^AF/_h  |   ·h
⇒ AF = ^h/_{tan α} = ^2,57cm/_{tan 40°},     AF = 3,05cm

Die Grundseite a = AF + FG +GB, AF = GB
⇒ a = 2AF + FG   |   −2AF
⇒ FG = a − 2AF = 9,2cm − 2·3,05cm,     FG = 3,1cm
Mit FG = DC = c, haben wir c = 3,1cm

Flächeninhalt A = ^{(a + c)·h}/₂ = ^{(9.2cm + 3,1cm)·2,57cm}/₂ = 15,80cm²     A = 15,80cm²


Gegeben: a = 5,1cm, h = 3,2cm, γ = 108°
Gesucht: b = d, c, α = β und A

γ = δ = 108°
360° = α + β + γ + δ = 2α + 2γ = 2α + 216  |  −216°
144° = 2α   |  ÷2
⇒ α = β = 72°

sin α = ^h/_d   |  ·(−1)
⇒ ¹/_{sin α} = ^d/_h   |  ·h
⇒ d = ^h/_{sin α} = ^3,2cm/_{sin 72°} = 3,36cm
d = b = 3,36cm

AF = GB
cos 72° = ^AF/_d   |  ·d
⇒ AF = d·cos 72° = 3,36cm·cos 72°
⇒ AF = GB = 1,04cm

FG = c, a = 5,1cm
5,1cm = 1,04cm + FG + 1,04cm = 2,08 + FG   |  −2,08      ⇒ c = FG = 3,02cm

A = ^{(a + c)·h}/₂ = ^{(5,1cm + 3,02cm)·3,2cm}/₂    ⇒   A ≈ 13cm²


Gegeben: b = 7,5cm, c = 3,4cm, h = 5,0cm
Gesucht: d, a, α = β γ = δ, AF = GB und A



Merke

Wie bei a) und b) sind immer wieder die Gleiche Formeln benutzt um die fehlende Größe zu berechnen!

So:
- d = b = 7,5cm
- sin α = ^h/_d = ^5,0cm/_7,5 = 0,66
  ⇒ sin α = 0,66   |   ÷sin
  ⇒ α = β ≈ 41,3°
- cos α = ^AF/_d ⇔ AF = d·cos 41,3° = 7,5cm·cos 41,3°
  ⇒ AF = GB = 5,63cm und FG = c = 3,4cm
- 360° = α + β + γ + δ = 2α + 2γ = 82,6° + 2γ
⇒ 360° = 82,6° + 2γ   |   −82,6°
  2γ = 277,4   |   ÷2
  ⇒ γ = δ = 138,7°
- a = AF + FG + GB = 5,63cm + 3,4cm + 5,63cm
  ⇒ a =14,66cm
- A = ^{(a + c)·h}/₂ = A = ^{(14,66cm + 3,4cm)·7,5cm}/₂
  ⇒ A = 67,72cm²


4. ... Still to come ...

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To Solution 4►►

Lösung 4:

Symmetrischer Drache

a)

• Länge der Diagonale e im Verhältnis 1:2

Gegeben: α = 39°, f = 7cm
Gesucht: e, A
f teilt die Diagonale e im Verhältnis 1:2 bedeudet daß die ganze Strecke e gleich 3 ist und f teilt e genau an der Teile 1 (also ^e/₃) (Siehe Abb.1).



Abb.1: Drache im Koordinaten System (Leicht erklärt)
so haben wir:
tan (^α/₂) = ^f/₂ ÷ ^e/₃   |   ·(−1)
   ⇒ ¹/_{tan (^α/2)} = ^e/₃ ÷ ^f/₂   |   ·^f/₂
   ⇒ ^e/₃ = ^f/₂ · ¹/_{tan (^α/2)}   |   ·3
   ⇒ e = 3( ^f/₂ · ¹/_{tan (^α/2)})

Bekannte Werte einsetzen: ^f/₂ = ⁷/₂ = 3,5cm; α = 39° und ^α/₂ = 19,5°
   ⇒ e = 3(3,5cm · 2,82) = 29,66 cm


• Flächeninhalt A des Drachenvierecks
A = ^(e·f)/₂ = ^{(29,66cm · 7cm)}/₂ = 103,81 cm    ⇒ A = 103,8 cm²

b)
Gesucht: β, γ, a und b

• DM = MB = 3,5cm, ^α/₂ = 19,5°
tan 19,5° = ^3,5cm/_AM
⇒ AM = ^3,5cm/_0,35,    ⇒ AM = 9,88cm
cos 19,5° = ^AM/_a = ^9,88cm/_a   |   ·a
   ⇒ a = 10,48cm

• In dem Dreieck AMB (β₁), ist der gesamte Winkel: ^α/₂ + 90° + ^β₁/₂ = 180°
^α/₂ = 19,5°
⇒ 19,5° + 90° + ^β₁/₂ = 180°   |   −109,5°
⇒ ^β₁/₂ = 70,5°    ⇒ ^β₁/₂ = 70,5°

• In dem Dreieck BMC, ist ^f/₂ = 3,5cm, und ^2e/₃ = 19,77cm die Länge MC:
  so, tan (^γ/₂) = ^3,5cm/_19,77cm = 0,177   |   ÷tan
  ⇒ ^γ/₂ = tan⁻¹ (0.177) = 10,03° und γ = 20,06°
  
  wieder in BMC: β₂
  ^γ/₂ + 90° + ^β₂/₂ = 180° ⇔ 10,03° + 90° + ^β₂/₂ = 180°   |   −100,03°
  ⇒ ^β₂/₂ ≈ 80°
  
  β = ^β₁/₂ + ^β₂/₂ = 150.5° ⇒ β = θ = 150,5° (θ gegen winkel zu β)
  
  der gesamte Winkel: α + β + γ + θ = 360°
  
  
• In dem Dreieck BMC, sind die Winkeln: ^β₂/₂ = 80°, ^γ/₂ = 10,03° und DM = MB = 3,5cm
sin (10,03°) = ^3,5cm/_b   |   ·(−1)
⇒ ¹/_{sin (10,03°)} = ^b/_3,5cm   |   ·3,5cm
⇒ b = ^3,5cm/_{sin (10,03°)} ≈ 20,1cm
⇒ b ≈ 20,1cm

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To Solution 5►►

Lösung 5:

... Still to come ...

Lösung zur Trigonometrie (1)

Lösung 2:

1. Dreieck aus dem Parallelogramm ABCD

2. Dreieck aus dem Parallelogramm ABCD

Gegeben: a = 4,1 cm und b = 3,4 cm
Gesucht: h₁, h₂ und A.

a) α = 42°

• Aus dem 2. Dreieck:
  sin α = ^h₁/_b  | ·b
  ⇒ h₁ = b·sin α = 3,4 cm · sin 42° ≈ 2,3 cm,        h₁ = 2,3 cm
• Nach dem 1. Dreieck:
  sin α = ^h₂/_a  | ·a
  ⇒ h₂ = a·sin α = 4,1 cm · sin 42° = 2,74 cm,      h₂ = 2,74 cm
• Der Flächeninhalt eines Parallelogramms ist eine Mischung aus Dreieck und Rechteck. Es gilt denn A = g · h, mit g die Grundseite des Rechtecks und h die Höhe des Dreiecks.
  In unserem Fall, g = a und h = h₁ (siehe Abbildung A)
  ⇒ A = a · h₁ = 4,1 cm · 2,3 cm
  ⇒ A = 9,43 cm²

Abbildung A: Flächeninhalt des Parallelogramms

b) α = 115°
Hier brauchst du nur noch α = 115° überall in a) einsetzen !!!
so bekommst du:

• h₁ = 3,4 cm · sin 115° = 3,08 cm
h₁ = 3,08 cm
• h₂ = 4,1 cm · sin 115° = 3,71 cm
h₂ = 3,71 cm
• A = a · h₁ = 4,1 cm · 3,08 cm = 12,63 cm²
A = 12,63 cm²

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To Solution 3►►

Konstante Steigung (Steigung Lineare Funktionen)

Lineare Funktion

• Eine Gleichung der Form: y = mx + b, mit x ∈ ℝ und m, b Konstante.
• Der Graph ist eine Gerade (Siehe Abb.).
• Definitionsbereich: D = ℝ ( ]−∞, +∞[ ).
• Der Achsenabschnitt: Schnitt mit y-Achse ist b.
• Die Steigung m der Abgebildeten Funktion lässt sich durch ∆x und ∆y ableiten oder bestimmen! (mit Steigunsformel und aus dem Steigungsdreieck).

Abbildung1: Steigung Lineare Funktion

So wird's gerechnet:
- ∆x: x-Angaben; (x₁ und x₂)
- ∆y: y-Angaben; (y₁ und y₂)
- Y-Achse: y = 0 (in diesem Fall)

m =

∆y -------- = ∆x

y2 − y1 ---------- = x2 − 11

1 − 4 ---------- − 1 + 4

= − 1

Die Steigung der Funktion lautet: m = − 1
Die Funktionsgleichung: y = − x

Übung zur Trigonometrie

being processed...

Aufgaben

Lösungen

Lösung 1:

1. Gegeben: a = b = 5,9cm, α = β = 32°
Gesucht: c, h, γ und A
Bekannt: α + β + γ = 180 ° ⇔ 32° + 32° + γ = 180° ⇒ γ = 180° - 64° = 116°
cosβ = ^BD/_a ⇔ BD = 5,9cm · cos32° ⇒ BD = AD = 5,0cm
es folgt c= 10cm
Flächeninhalt A des Dreiecks ADC + BDC = 2(¹/₂·AD·h), wo sinα = ^h/_a und h = a·sinα:
A = 2(¹/₂·5·5,9·sinα) = 15,6cm², A = 15,6cm²


2. Gegeben: a = b = 4,5dm, γ = 98°
Gesucht: c, h, α = β und A
Bekannt: α + β + γ = 180 °, AD = DB
⇔ 2α + γ = 180° | − 98°
⇒ 2α = 82 | ÷ 2
⇒ α = β = 41°
cosα = ^AD/_b und AD = bcosα = 4,5dm · cos41*deg; ≈ 3,4
AD = DB ≈ 3,4cm, und c = 6,8dm
Flächeninhalt A des Dreiecks ADC + BDC = 2(¹/₂·AD·h), wo sinα = ^h/_a und h = a·sinα:
A = 2(¹/₂·3,4·4,5·sin41°) = 10,03dm², A = 10,03dm²


3. Gegeben: a = b = 65,4m, c = 54,7m
Gesucht: h, α = β, und γ
Bekannt: α + β + γ = 180 °, c= AD + DB, mit AD = BD ⇒ c = 2AD und AD = ^c/₂
AD = BD = 27,35m
cosβ = ^BD/_a = ^27,35/_65,4
⇒ cosβ ≈ 0,42  | ÷cos
⇒ β = cos^-1(0,42) = 65,2°, α = β = 65,2°
wir wissen dass α + β + γ = 180 ° (α und β einsetzen um γ zu rechen)
⇒ 2(65,2) + γ = 180 °  | −130,4
γ = 49,6°

Wir brauchen die Höhe h um den Flächeninhalt ein Rechtwinkliges Dreieck zu rechen.
Nach der Formel: A = ¹/₂·AD·h
In den Dreiecken ADC und BDC ist die Höhe h gleich. Man kann denn den Flächeninhalt eines Dreiecks rechnen und mal 2 multiplizieren
sinα = ^h/_b | ·b
⇒ h = b·sinα, h = 65.4m · sin 65,2° = 59,37m
Der Flächeninhalt ist denn: A = 2(¹/₂·27,35·59,73) = 816,8m²

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To Solution 2►►

Trigonometrie

Rechtwinkliges Dreieck: Sinus, Kosinus, Tangens, Kotangens

Wichtig

Die Seitenverhältnisse in einem rechtwinkligen Dreieck stehen in Beziehung zu den Winkeln, so dass man die Winkel über die Seitenverhältnisse bestimmer kann.

Winkelfunktionen im rechtwinkligen Dreieck

Abbildung1: Rechtwinkliges Dreieck

Gemäß der Abbildung:

> sin α =

Gegenkathete von α -------------------------- Hypothenuse

=

a --- c

=

cos β

> cos α =

Ankathete von α -------------------------- Hypothenuse

=

b --- c

=

sin β

> tan α =

Gegenkathete von α ---------------------------- Ankathete von α

=

a --- b

=

cot β

> cot α =

Ankathete von α ------------------------ Gegenkathete von α

=

b --- c

=

tan α

Umrechnungen Grad und Bogenmaß

Wichtig

Es besteht ein Zusammenhang zwischen einem Winkel in Grad und der Länge des dazugehörigen Bogenmaßes.

α = 180° --------- • π x ⇔ x = π --------- • 180° α

Grad α	10°	30°	45°	90°	120°	180°	270°	360°
Bogenmaß x	π ----- 18	π ----- 6	π ----- 4	π ----- 2	2π ----- 3	π	3π ----- 2	2π

Wichtig

Am Einheitskreis (Kreis mit Radius 1) lassen sich die Winkelfunktionen anschaulich darstellen.

Bemerkenswerte von Sinus und Kosinus, Tangens und Kotangens

Wert(Grad) ► Winkel(α) ▼	0°	30°	45°	60°	90°
sin	0	0.5	0.707	0.86	1
cos	1	0.86	0.707	0.5	0
tan	0	0.58	1	1.72	Unmgl.
cot	Unmgl.	1.72	1	0.58	0

Zur Übung

Solutions to Exercises on Vertex

1. Find (i) the vertex and (ii) the equation of the line of symetry of each of the following quadratic graphs.


1. y = 1(x -2)² + 3
   according to the formula, the quadratic equation is in the Vertex form:
   
   (i) its vertex is v(h, k), with h = -(-2) and k = 3 ==> v(2, 3)
   1 > 0, parabola opens upwards => Minimum
   
   (ii) its line of symetry is x = 2 (see grah)
   
   
   



The same method can be applied to b., c., d., e. and f.



7. y = 1(x -3)² + c
   
   (i) Vertex v(3, c)
   with c undefined constant, it acts like a parameter!!!
   1 > 0, parabola opens upwards => Minimum
   
   (ii) The line of symetry of the equation is x = c, c can take any value!


8. y = 1(x -p)² + q
   
   (i) h = p and k = q ==> vertex v(p, q)
   p, q constant values;
   1 > 0, parabola opens upwards => Minimum
   
   (ii) Line of symetry x = p, p constant.


9. y = (ax + b)² + c
   first bring the equation to a vertex form (put a as factor as in the vertex form): ==> y = a(x + ^b/_a) + c (vertex form)
   
   (i) h = −^b/_a, k = c, and the vertex: v(−^b/_a, c), a, b and c constant.
   a > 0, parabola opens upwards => Minimum
   
   (ii) Line of symetry x = −^b/_a, a and b constant.



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2. Find (i) the least (or, if appropriate, the greatest) value of each of the following quadratic expressions and (ii) the value of x for which this occurs.




Core Lesson



• Vertex form of a quadratic function: f(x) = a(x - h)² + k.
• a determines how the parabola opens:
  - a > 0, it opens upwards;
  - a < 0, it opens downwards.
• h and k are the values of the vertex.
• The greatest or least value of a quadratic function equals k.
• The greatest or least value of a quadratic function occurs at x = h.
• Vertex: v(h, k)


1. y = 1(x + 2)² - 1
   (x + 2)² - 1 is smallest for x + 2 = 0
   x + 2 = 0 when x = −2
   ==> the function will have its smallest output when x = −2
   
   so let's find the output now:
   f(-2) = (-2 + 2)² - 1 = (0)² −1 = −1
   
   So the least value of y = 1(x + 2)² - 1 is k = −1 and it occurs at x = −2, and the parabola opens upwards for a = 1 > 0.


2. y = 1(x - 1)² + 2
a = 1 > 0, the parabola opens upwards
==> its least value is k = 2 and it occurs at x = 1.


3. y = 5 - 1(x + 3)²
or y = −1(x + 3)² + 5
a = −1 < 0, the parabola opens downwards
==> its greatest value is k = 5 and it occurs at x = −3.


4. y = 1(2x + 1)² − 7
by factorizing 2: 2(x + ¹/₂)² − 7
a = 2 > 0, the parabola opens upwards
==> its least value is k = −7 and it occurs at x = ⁻¹/₂.


5. y = 3 − 2(x − 4)² or y = − 2(x − 4)² + 3
a = −2 < 0, the parabola opens downwards
==> the least value of the function is k = 3 and it occurs at x = 4.


6. y = 1(x + p)² + q, with p and q constant.
a = 1 > 0, the parabola opens upwards
==> the greatest value of the function is k = q and it occurs at x = −p.


7. y = 1(x − p)² − q, with p and q constant.
a = 1 > 0, the parabola opens upwards
==> the greatest value of the function is k = −q and it occurs at x = p.


8. y = r −1(x − t)² or y = −1(x − t)² + r, with t and r constant.
a = −1 < 0, the parabola opens downwards
==> the least value of the function is k = r and it occurs at x = t.


9. y = c −1(ax + b)² or y = −1(ax + b)² + c, with a, b and c constant.
factorize by a: −a(x + ^b/_a) + c (vertex form)
−a < 0, the parabola opens downwards
==> the least value of the function is k = c and it occurs at x = ^−b/_a.



3. Solve the following quadratic equations. Leave surds in your answer.


1. (x − 3)² − 3 = 0   | +3 (on both side)
(x − 3)² = 3   | √ (square root it)
x − 3 = ± √3   | +3
==> x = 3 ± √3


2. (x + 2)² − 4 = 0   | +4
(x + 2)² = 4   | √ (square root it)
x + 2 = ± 2
==> x = − 2 ± √2


3. 2(x + 3)² = 5   | ÷2
(x + 3)² = ⁵/₂   | √
x + 3 = ± √⁵/₂   | −3
==> x = − 3 ± √⁵/₂


4. (3x − 7)² = 8   | √
3x − 7 = ± √8   | +7
3x = 7 ± 2√2   | ÷3
==> x = ⁷/₃ ± ^2√2/₃


5. (x + p)² − q = 0   | +q
(x + p)² = q   | √
x + p = ± √q   | −p
==> x = − p ± √q


6. a(x + b)² − c = 0   | +c
a(x + b)² = c   | ÷a
(x + b)² = ^c/_a   | √
x + b = ± √^c/_a   | −b
==> x = − b ± √^c/_a.



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4. Express the following in completed square form


1. x² + 2x +2 (by completing the square)
(x² + 2x + 1) + 2 − 1
==> (x + 1)² + 1


2. x² −8x − 3 (completing the square)
(x² − 8x + 16)² − 3 − 16
==> (x − 4)² − 19


3. x² + 3x − 7 (completing the square)
(x² + 3x + 2.25)² − 7 − 2.25
==> (x + 1.5)² − 9.25


4. 5 − 6x + x², better x² − 6x + 5
(x² − 6x + 9) + 5 − 9
==> (x − 3)² − 4 or (x − 3)² − 2²


5. x² + 14x + 49 (This is a completed square Form)
==> (x + 7)²


6. 2x² + 12x − 5   |   ÷2 (divide by 2 to bring the equation to a regular quadratic form)
x² + 6x − ⁵/₂ (complete the square)
(x² + 6x + 9) − ⁵/₂ − 9
==> (x + 3)² − ²³/₂


7. 3x² − 12x + 3   |   ÷3
x² − 4x + 1 (complete the square)
(x² − 4x + 4) + 1 − 4
==> (x − 2)² − 3


8. 7 − 8x − 4x²   |   ÷4
<==> −x − 2x + ⁷/₄   |   ×(−1)
<==> x + 2x −⁷/₄   |   (complete the square)
(x² + 2x + 1) −⁷/₄ − 1
==> (x + 1) −¹¹/₄


9. 2x² + 5x − 3   |   ÷2
x² + ⁵/₂ − ³/₂   |   (complete the square)
(x² + ⁵/₂ + 1.56) − ³/₂ − 1.56
==> (x + 1.25) − 3.06




5. Use the completed square form to factorize the following expressions
1. x² − 2x − 35   |   (complete the square)
   
(x² − 2x + 1) − 35 − 1
(x − 1)² − 36 <==> (x − 1)² − 6²   |   √ (square it)
±(x + 1) ± 6
==> (x + 1 − 6)(x + 1 + 6)
==> (x − 5)(x + 7)


2. x² − 14x − 176   |   (complete the square)
   
(x² − 14 + 49) − 176 − 49 <==> (x − 7)² − 225 <==> (x − 7)² − 15²  |   √
==> (x − 7 + 15)(x − 7 − 15)
==> (x + 8)(x − 22)


3. x² + 6x − 432   |   (complete the square)
   
(x² + 6x + 9) − 432 − 9 <==> (x + 3)² − 441
(x + 3)² − 21²   |   √
==> (x + 3 − 21)(x + 3 + 21)
==> (x − 18)(x + 24)


4. 6x − 5x − 6   |   ÷6
   
x² − ^5x/₆ − 1   |   (complete the square)
(x² − ^5x/₆ + 0.18) − 1 − 0.18
(x − 0.42)² − 1.18 <==> (x − 0.42)² − ±1.08   |   √
==> (x − 0.42 − 1.09)(x − 0.42 + 1.09)
==> (x − 1.5)(x + 0.67)


5. 14 + 45x − 14x² <==> − 14x² + 45x + 14   |   − ÷14
x² − ^45x/₁₄ − 1   |   (complete the square)
(x² − 3.2x + 1.6²) − 1 − 1.6²
(x − 1.6)² − 3.5 <==> (x − 1.6)² − 1.9²   |   √
==> (x − 1.6 − 1.9)(x − 1.6 + 1.9)
==> (x − 3.5)(x − 0.3)


6. 12x² + x − 6   |   ÷12
x² − ^x/₁₂−¹/₂  |   (complete the square)
(x² − ^x/₁₂ + 0.041²) − 0.5 − 0.041²
(x − 0.041)² − 0.5 <==> (x − 0.041)² ± 0.7²
==> (x − 0.041 − 0.7)(x − 0.041 + 0.7)
==> (x − 0.74)(x + 0.66)



6. See Exercise 2

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BIS Berlin Pure Mathematics 1

Exercise 4A

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