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Samenvatting Wiskunde B HAVO Hoofdstuk 11 Verbanden en functies 2020) $2.60   Add to cart

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Samenvatting Wiskunde B HAVO Hoofdstuk 11 Verbanden en functies 2020)

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Wiskunde B HAVO Hoofdstuk 11 Verbanden en functies 2020) PARAGRAAF 11.0 : VOORKENNIS PARAGRAAF 11.1 : STELSELS BIJ PARABOLEN PARAGRAAF 11.2 : (OMGEKEERD) EVENREDIG PARAGRAAF 11.3 : STANDAARDFUNCTIES PARAGRAAF 11.4 : WORTELFUNCTIES EN GEBROKEN FUNCTIES PARAGRAAF 11.5 : OPTIMALISERINGSPROBLEM...

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  • August 2, 2023
  • 18
  • 2020/2021
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Hoofdstuk 11 Verbanden en functies (H5 Wis B) Pagina 1 van 18



PARAGRAAF 11.0 : VOORKENNIS


LES 1 : STELSELS, FORMULES EN AFGELEIDE


VOORBEELD 1

Los op.
3𝑥 + 5𝑦 = −7
a. {
2𝑥 + 𝑦 = 0
2𝑥 + 5𝑦 = 38
b. {
𝑥 =𝑦+5


OPLOSSING 1

a.
3𝑥 + 5𝑦 = −7 1 3𝑥 + 5𝑦 = −7
(1) { | | geeft {
2𝑥 + 𝑦 = 0 5 10𝑥 + 5𝑦 = 0
__________ –
– 7𝑥 + 0 = – 7
𝑥 = 1
(2) 3 · 1 + 5𝑦 = – 7
5𝑦 = – 10
𝑦 = –2
dus oplossing is (1, −2)



b.
2𝑥 + 5𝑦 = 38
(1) {  x = y + 5 invullen (substitutie)
𝑥 =𝑦+5

2(𝑦 + 5) + 5𝑦 = 38
2𝑦 + 10 + 5𝑦 = 38
7𝑦 + 10 = 38
7𝑦 = 28
𝑦 = 4


(2) dus 𝑥 = 4 + 5 = 9
Snijpunt (9,4)

,Hoofdstuk 11 Verbanden en functies (H5 Wis B) Pagina 2 van 18



LES 2 : KWADRATISCHE FORMULES

Er zijn 3 notaties voor kwadratische vergelijkingen :


(1) 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
−𝑏
Top berekenen door: xtop = 2𝑎
of afgeleide = 0


(2) 𝑦 = 𝑎(𝑥 – 𝑑)(𝑥 – 𝑒) snijdt de x-as in de punten (𝑑, 0) en (𝑒, 0).
𝑑+𝑒
De top ligt precies tussen de snijpunten met de x-as → xtop = 2


(3) 𝑦 = 𝑎(𝑥 – 𝑝)2 + 𝑞
𝑇𝑜𝑝 = (𝑝, 𝑞)


VOORBEELD 1

Stel de formules op van de volgende parabolen.
a. met top(3,10) en door het punt 𝐴(1,4).
b. door de punten 𝐴(−3,0), 𝐵(7,0) 𝑒𝑛 𝐶(6,4).


OPLOSSING 1

a. 𝑝 = 3 𝑒𝑛 𝑞 = 10 → 𝑦 = 𝑎(𝑥 – 3)2 + 10
(1,4) → 4 = 𝑎(1 – 3)2 + 10
4 = 𝑎 ∙ 4 + 10
−6 = 𝑎∙4
1
𝑎 = −1 2
1
𝑦 = −1 2 (𝑥 – 3)2 + 10
1
𝑦 = −1 2 (𝑥 2 – 6𝑥 + 9) + 10
1 1
𝑦 = −1 𝑥 2 + 9𝑥 + 23
2 2


b. De snijpunten met de x-as zijn: 𝐴(−3,0) 𝑒𝑛 𝐵(7,0) → 𝑑 = −3 𝑒𝑛 𝑒 = 7
In 𝑦 = 𝑎(𝑥 – 𝑑)(𝑥 – 𝑒) invullen geeft 𝑦 = 𝑎(𝑥 + 3)(𝑥 – 7)
𝐶(6,4) → 4 = 𝑎(6 + 3)(6 – 7)
4 = 𝑎 ∙ −9
𝑎 = − 49
Dus 𝑦 = − 49 (𝑥 + 3)(𝑥 – 7) = − 49 (𝑥 2 – 4𝑥 – 21)
7 1
𝑦 = − 49 𝑥 2 + 1 9 𝑥 + 9 3

, Hoofdstuk 11 Verbanden en functies (H5 Wis B) Pagina 3 van 18



LES 3 : DE AFGELEIDE FUNCTIE



REGELS DIFFERENTIËREN
(1) Hoofdregel: 𝑓(𝑥) = 𝑎 ∙ 𝑥 𝑛 → 𝑓 ‘ (𝑥) = 𝑛 ∙ 𝑎 ∙ 𝑥 𝑛−1
(2) Kettingregel: 𝑓′(𝑥) = 𝑓′(𝑢) ∙ 𝑢′(𝑥))




VOORBEELD 1

Differentieer de functie 𝑓(𝑥) = (10 − 4𝑥)7




OPLOSSING 1

(1) Neem 𝑢 = 10 − 4𝑥 → 𝑢′ = −4
Dan is 𝑓(𝑢) = 𝑢7 → 𝑓′(𝑢) = 7𝑢6


(2) 𝑓 ′ (𝑥) = 𝑓 ′ (𝑢) ∙ 𝑢′ (𝑥) = −4 ∙ 7𝑢6 = −4 ∙ 7(10 − 4𝑥)6 =
𝑓 ′ (𝑥) = −28(10 − 4𝑥)6




SNELLE MANIER

Het kan ook sneller zonder de u. De macht ervoor, de macht eentje lager en
vermenigvuldigen met de afgeleide van wat tussen haakjes staat.
𝑓 ′ (𝑥) = −4 ∙ 7(10 − 4𝑥)6 = −28(10 − 4𝑥)6

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