Lösungen: zu 1.
P(2;1) und Q(5;4)
I. 1 = 2² + 2p + q
II. 4 = 5² + 5p + q
I. q = 1 – 4 –
2p q = -3 - 2p
II. q = 4 – 25
-5p q = -21 - 5p
I = II: -3 – 2p = -21 –
5p →
p = -6 →
q = 9
→
y = x² - 6x + 9
→
S(-d;e) y = x² -6x + 9 + 3² - 3²
→ y = (x –
3)²
→
S(3;0)
zu 2.
P(1;0) und Q(4;-3)
I. 0 = 1² + 1p + q
II. -3 = 4² + 4p + q
I. q = 0 – 1² - 1p q = -1 - 1p
II. q = -3 – 4² - 4p q = -19 - 4p
→
I = II: -1 – p = -19 – 4p
→ p
= -6 →
q = 5
→
y = x² - 6x + 5
→
S(-d;e) y = x² -6x + 5 + 3² - 3²
à
y = (x – 3)² - 4
→
S(3;-4)
0 = x² -6x + 5
→
x1/2 =
→ x1 = 3 + 2 = 5
→ x2 = 3 - 2 = 1
a) S(3;-4)
b) -4 y
∞
c) -∞
x
∞
d) -∞
x
3 monoton fallend
3
x
∞ monoton steigend
e) x1=
5 und
x2=
1
f) ymin = -4
g) Py(0;5)
zur Kontrolle:

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