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Samenvatting 4VWO getal en ruimte wiskunde B hoofdstuk 1 en 2
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samenvatting wiskunde b getal en ruimte hoofdstuk 1 en 2
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Wiskunde 4VWO hoofdstuk 1 functies en grafiekenvergelijkingen
Evenwijdige lijnen hebben hetzelfde richtingscoëfficiënt.
Δ𝑦
RC= Δ𝑥
Bij 𝑦 = 𝑎𝑥 + 𝑏 is y een lineaire functie van X
Oplossen van tweetermen
5x2 – 7x = 0 3x2= 30
X(5x-7)= 0 X2=10
X= 0 v 5x-7=0 X= √10 v x= −√10
X=0 v 5x=7 [ : 5]
X=0 v x = 1,4 KWADRAAT NOOIT NEGATIEF!!
Oplossen van drietermen
x2-x -6=0 X2-7x+2=0 ABC-formule
(x+2)(x-3)= 0 (x+3,5)2- (3,5)2 +2=0 D= b2- 4ac
X= -2 v x= 3 (x+3,5)2- 12,25 +2=0 D< 0 o oplossingen
(x+3,5)2 = 10, 25 D=0 1 oplossingen
Iets x iets =0 x+3,5= √10, 25 v x+3,5= −√10, 25 D>0 2 oplossingen
anders niet −𝒃 + √𝑫 −𝒃 − √𝑫
deze manier 𝒙= 𝒗𝒙=
𝟐𝒂 𝟐𝒂
gebruiken!
𝒇𝒑 = 𝟐𝒙2-10x+p
D> 0 delen door negatief getal→ teken klapt om
D=100-8p
100-8p>0
P<12 1/3
𝒇𝒑(𝒙) = 𝟒𝒙2+px +3 minimum=1 p=?
1
𝑥𝑡𝑜𝑝 = − 4 𝑝
1 1
𝑦𝑡𝑜𝑝(− 4 𝑝) = − 8 𝑝2+3 p2 heeft 2 uitkomsten (pos en neg)!
1
− 8 𝑝2+3=1
𝑝 = 4 𝑣𝑝 = −4
Bereik en domein
Bereik [B]= alle mogelijk y-waarden
Domein [D] = alle mogelijke x-waarden
<,→>=ℝ
Extreme waarden= top (neg/pos)
−𝒃
𝒙𝒕𝒐𝒑 = 𝟐𝒂
𝐷𝑓 = [0,8] 𝐵𝑓 =? 𝑓(𝑥) = 0,4𝑥2-2x +2
𝑓(0) = 2 𝑓(8) = 5,2
−𝑏 −2,8
𝑥𝑡𝑜𝑝 = = = 3,5
2𝑎 0,8
𝑦𝑡𝑜𝑝(3,5) = −2,9
Bf= [-2,9;5,2]
, Kromme door de top
𝒇𝒑(𝒙) = 𝟐𝒙 + 𝒑𝒙 − 𝟑
−𝑝
1. 𝑥 𝑡𝑜𝑝 = 4
2. herschrijven tot p= -4x
3. Invullen 𝑓𝑝(−4𝑥) = −2𝑥2-3
4. y= -2x2-3 (andere naam dan f)
Bereken voor welke p de vergelijking px2+2x+5=0 twee oplossingen heeft
D>0 dus 1 oplossing, p kan geen 0 zijn want dan is het geen parabool meer
D=4-20p
4-20p>0 Grafisch rekenmachine noteren
P<0,2 1) Invoer
P<0 v 0<p<0,2 2) Window
3) Optie
Grafische numeriek oplossen 4) Antwoord
5) (schets)
6) (finale antwoord)
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