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# Hiperboliese funksies en grafieke

## Inleiding

In graag 10 het jy verskillende grafieke se vorms bestudeer. In hierdie hoofstuk sal jy leer van grafieke van funksies.

## Funksies in die Vorm y=ax+p+qy=ax+p+q

Hierdie vorm van die hiperboliese funksie is effens meer kompleks as die vorms wat in graad 10 teëgekom is.

### Ondersoek: Funksies van die Vorm y=ax+p+qy=ax+p+q

1. Op dieselfde assestelsel, teken die volgende grafieke:
1. a(x)=-2x+1+1a(x)=-2x+1+1
2. b(x)=-1x+1+1b(x)=-1x+1+1
3. c(x)=0x+1+1c(x)=0x+1+1
4. d(x)=1x+1+1d(x)=1x+1+1
5. e(x)=2x+1+1e(x)=2x+1+1
Gebruik die resultate om die effek af te lei van Use your results to deduce the effect of aa.
2. Op dieselfde assestelsel, teken die volgende grafieke:
1. f(x)=1x-2+1f(x)=1x-2+1
2. g(x)=1x-1+1g(x)=1x-1+1
3. h(x)=1x+0+1h(x)=1x+0+1
4. j(x)=1x+1+1j(x)=1x+1+1
5. k(x)=1x+2+1k(x)=1x+2+1
Gebruik jou resultate om die effekte af te lei van pp.
3. Deur die algemene metode van die bogenoemde aktiwiteite, kies jou eie waardes van aa en pp om 5 verskillende grafieke te teken van y=ax+p+qy=ax+p+q om die effekte van qq af te lei.

Jy behoort te gevind het dat die teken van aa beïnvloed of die grafiek in die eerste en derde of in die tweede en vierde kwadrant van die Cartesiese vlak is.

Jy sou ook gevind het dat die waarde vand pp beïnvloed of die xx-afsnit negatief (p>0p>0) of positief(p<0p<0) is.

Jy behoort ook te gevind het dat die waarde van qq beïnvloed of die grafiek bo diexx-as (q>0q>0) of onder die xx-as (q<0q<0) lê.

Hierdie verskillende eienskappe word opgesom in tabel 1. Die asse van simmetrie vir elke grafiek word vertoon as ‘n stippellyn.

 p < 0 p < 0 p > 0 p > 0 a > 0 a > 0 a < 0 a < 0 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

### Gebied en Terrein

Vir y=ax+p+qy=ax+p+q, is die funksie ongedefinieerd vir x=-px=-p. Die gebied is daarom {x:xR,x-p}{x:xR,x-p}.

Ons sien dat y=ax+p+qy=ax+p+q kan herskryf word as:

y = a x + p + q y - q = a x + p As x - p dan is : ( y - q ) ( x + p ) = a x + p = a y - q y = a x + p + q y - q = a x + p As x - p dan is : ( y - q ) ( x + p ) = a x + p = a y - q
(1)

Dit wys dat die funksie ongedefinieerd is by y=qy=q. Die terrein van f(x)=ax+p+qf(x)=ax+p+q is daarom {f(x):f(x)R,f(x)q{f(x):f(x)R,f(x)q.

Byvoorbeeld, die gebied van g(x)=2x+1+2g(x)=2x+1+2 is {x:xR,x-1}{x:xR,x-1} want g(x)g(x) is ongedefinieerd by x=-1x=-1.

y = 2 x + 1 + 2 ( y - 2 ) = 2 x + 1 ( y - 2 ) ( x + 1 ) = 2 ( x + 1 ) = 2 y - 2 y = 2 x + 1 + 2 ( y - 2 ) = 2 x + 1 ( y - 2 ) ( x + 1 ) = 2 ( x + 1 ) = 2 y - 2
(2)

Ons kan sien dat g(x)g(x) is ongedefinieerd by y=2y=2. Daarom is die gebied {g(x):g(x)(-,2)(2,)}{g(x):g(x)(-,2)(2,)}.

#### Gebied en Terrein

1. Bepaal die terrein van y=1x+1y=1x+1.
2. Gegewe:f(x)=8x-8+4f(x)=8x-8+4. Write down the domain of ff.
3. Bepaal die gebied van y=-8x+1+3y=-8x+1+3

### Afsnitte

Vir funksies van die vorm, y=ax+p+qy=ax+p+q, word die afsnitte met die xx en yy assebereken deur x=0x=0 te stel vir die yy-afsnit en deur y=0y=0 te stel vir die xx-afsnit.

The yy-intercept is calculated as follows:

y = a x + p + q y i n t = a 0 + p + q = a p + q y = a x + p + q y i n t = a 0 + p + q = a p + q
(3)

Byvoorbeeld, die yy-afsnit van g(x)=2x+1+2g(x)=2x+1+2 word verkry deur x=0x=0 te stel, wat lewer:

y = 2 x + 1 + 2 y i n t = 2 0 + 1 + 2 = 2 1 + 2 = 2 + 2 = 4 y = 2 x + 1 + 2 y i n t = 2 0 + 1 + 2 = 2 1 + 2 = 2 + 2 = 4
(4)

Die xx-afsnitte word bereken deur y=0y=0 te stel as volg:

y = a x + p + q 0 = a x i n t + p + q a x i n t + p = - q a = - q ( x i n t + p ) x i n t + p = a - q x i n t = a - q - p y = a x + p + q 0 = a x i n t + p + q a x i n t + p = - q a = - q ( x i n t + p ) x i n t + p = a - q x i n t = a - q - p
(5)

Byvoorbeeld, die xx-afsnit van g(x)=2x+1+2g(x)=2x+1+2 word gegee deur x=0x=0 te stel om die volgende te kry:

y = 2 x + 1 + 2 0 = 2 x i n t + 1 + 2 - 2 = 2 x i n t + 1 - 2 ( x i n t + 1 ) = 2 x i n t + 1 = 2 - 2 x i n t = - 1 - 1 x i n t = - 2 y = 2 x + 1 + 2 0 = 2 x i n t + 1 + 2 - 2 = 2 x i n t + 1 - 2 ( x i n t + 1 ) = 2 x i n t + 1 = 2 - 2 x i n t = - 1 - 1 x i n t = - 2
(6)

#### Afsnitte

1. Gegewe:h(x)=1x+4-2h(x)=1x+4-2. Bepaal die koördinate van die afsnitte van hh met die x- en y-asse.
2. Bepaal die x-afsnit van die grafiek van y=5x+2y=5x+2. Hoekom is daar geen y-afsnit vir hierdie funksie nie?

### Asimptote

Daar is twee asimptote vir funksies van die vorm y=ax+p+qy=ax+p+q. Hulle word bepaal deur die gebied en terrein te ondersoek.

Ons het gesien dat die funksie ongedefinieerd was by x=-px=-p en vir y=qy=q. Daarom is die asimptote x=-px=-p en y=qy=q.

Byvoorbeeld, die gebied van g(x)=2x+1+2g(x)=2x+1+2 is {x:xR,x-1}{x:xR,x-1} because g(x)g(x) is ongedefinieerd by x=-1x=-1. Ons sien ook dat g(x)g(x) is ongedefinieerd by y=2y=2. Daarom is die terrein {g(x):g(x)(-,2)(2,)}{g(x):g(x)(-,2)(2,)}.

Hieruit kan ons aflei dat die asimptote lê by x=-1x=-1 en y=2y=2.

#### Asimptote

1. Gegewe:h(x)=1x+4-2h(x)=1x+4-2. Bepaal die vergelykings van die asimptote van hh.
2. Skryf die vergelyking neer van die vertikale asimptoot van die funksie y=1x-1y=1x-1.

### Teken Grafieke van die Vormf(x)=ax+p+qf(x)=ax+p+q

Ten einde grafieke te teken van funksies van die vorm, f(x)=ax+p+qf(x)=ax+p+q, moet ons vier eienskappebepaal met berekeninge:

1. gebied en terrein
2. asimptote
3. yy-afsnit
4. xx-afsnit

Byvoorbeeld, teken die grafiek van g(x)=2x+1+2g(x)=2x+1+2. Dui die afsnitte en asimptote aan.

Ons het bepaal dat die gebied is {x:xR,x-1}{x:xR,x-1} en die terrein is {g(x):g(x)(-,2)(2,)}{g(x):g(x)(-,2)(2,)}. Daarom is die asimptote by x=-1x=-1 en y=2y=2.

Die yy-intercept is yint=4yint=4 en die xx-afsnit is xint=-2xint=-2.

#### Grafieke

1. Teken die grafiek van y=1x+2y=1x+2. Dui die horisontale asimptoot aan.
2. Gegewe:h(x)=1x+4-2h(x)=1x+4-2. Teken die grafiek van hh en dui duidelik die asimptote en ALLE afsnitte met die asse.
3. Teken die grafiek van y=1xy=1x en y=-8x+1+3y=-8x+1+3 op die selfdeassestelsel.
4. Teken die grafiek van y=5x-2,5+2y=5x-2,5+2. Verduidelik jou metode.
5. Teken die grafiek van die funksie gedefinieer deur y=8x-8+4y=8x-8+4. Dui die asimptote en die afsnitte met die asse aan.

## Einde van die Hoofstuk Oefeninge

1. Teken die grafeik van die hiperbool gedefinieer deur y=2xy=2x vir -4x4-4x4. Veronderstel die hiperbool word geskuif met 3 eenhede na regs en 1 eenheid af. Wat is die nuwe vergelyking nou?
2. Gebaseer op die grafiek van y=1xy=1x, bepaal die vergelyking van grafiek met asimptote y=2y=2 en x=1x=1 wat deur die punt (2; 3) gaan.

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