Odpowiedź:
1.
y = 3x² - 12x
a = 3 , b = - 12 , c = 0
Δ = b² - 4ac = (- 12)² - 4 * 3 * 0 = 144
W - współrzędne wierzchołka paraboli = (p , q)
p = - b/2a = 12/6 = 2
q = - Δ/4a = - 144/12 = - 12
W = ( 2 , - 12)
2.
y = 2x² + 20x + 49
a = 2 , b = 20 , c = 49
Δ = b² - 4ac = 20² - 4 * 2 * 49 = 400 - 392 = 8
p = - b/2a = - 20/4 = - 5
q = - Δ/4a = - 8/8 = - 1
W = ( - 5 , - 1 )
3.
y = - 4x² + 4x + 1
a = - 4 , b = 4 , c = 1
Δ = b² - 4ac = 4² - 4 * ²(- 4) * 1 = 16 + 16 = 32
p = - b/2a = - 4/(- 8) = 4/8 = 1/2
q = - Δ/4a = - 32/(- 16) = 32/16 = 2
W = ( 1/2 , 2 )
[tex]y = 3 {x}^{2} - 12 \\ ∆ = {0}^{2} - 4 \times 3 \times ( - 12) = 144 \\ p = - \frac{b}{2a} = - \frac{0}{6} = 0 \\ q = - \frac{∆}{4a} = - \frac{144}{3 \times 4} = - \frac{144}{12} = - 12[/tex]
W(0;-12)
[tex]y = 2 {x}^{2} + 20x + 49 \\ ∆ = {20}^{2} - 4 \times 2 \times 49 = 400 - 392 = 8 \\ p = - \frac{b}{2a} = - \frac{20}{2 \times 2} = - \frac{20}{4} = - 5 \\ q = - \frac{∆}{4a} = - \frac{8}{4 \times 2} = - \frac{8}{8} = - 1[/tex]
W(-5;-1)
[tex]y = - 4 {x}^{2} + 4x + 1 \\ ∆ = {4}^{2} - 4 \times ( - 4 )\times 1 = 16 + 16 = 32 \\ p = - \frac{4}{2 \times ( - 4)} = - \frac{4}{ - 8} = \frac{1}{2} \\ q = - \frac{∆}{4a} = - \frac{32}{4 \times ( - 4) } = \frac{32}{16} = 2[/tex]
W(1/2;2)
