Bardzo proszę o pomoc daje naj❣️

Odpowiedź:
Δ=b² - 4ac , Δ- wyróżnik funkcji kwadratowej
W=(p,q) - wierzchołek paraboli gdzie
p= -b/2a oraz q= -Δ/4a
zad.5
a) y=x²+4x-3
a=1,b=4,c= -3
Δ=4²-4×1×(-3)=16+12=28
p= -4/2 = -2
q= -28/4 = -7
W= (-2,-7)
b)y= -x²+6x-1
a= -1,b=6,c= -1
Δ=6²-4×(-1)×(-1)=36-4=32
p=-6/-2 =3
q=-32/-4 = 8
W= (3,8)
c)y=2x²+4x-5
a=2,b=4,c=-5
Δ=4²-4×2×(-5)=16+40=56
p=-4/4= -1
q=-56/8= -7
W=(-1,-7)
d)y= -3x²+6x-2
a= -3,b=6, c= -2
Δ=6²-4×(-3)×(-2)=36-24=12
p=-6/-6=1
q=-12/-12 = 1
W=(1,1)
zad.1
a) f(x)= -3x² + 6x - 4
f(2)= -3×2²+6×2-4= -12 +12 -4 = - 4
f(-1)= -3×(1)²+6×(-1)-4 = -3 - 6 - 4 = - 13
b) f(x)= 2x² - 3x - 5
f(2)=2×2² - 3×2 - 5 = 8 - 6 - 5 = - 3
f(-1)= 2×(-1)² - 3×(-1) - 5 = 2 + 3 - 5 = 0
5.
[tex]a)\\y = x^{2}+4x-3\\\\a = 1, \ b = 4, \ c = -3\\\\\Delta = b^{2}-4ac = 4^{2}-4\cdot1\cdot(-3) = 16+12 = 28\\\\W = (p, q)\\\\p = \frac{-b}{2a} = \frac{-4}{2} = -2, \ \ \ q = \frac{-\Delta}{4a} = \frac{-28}{4} = -7\\\\W = (-2, -7)[/tex]
[tex]b)\\y = -x^{2}+6x-1\\\\a = -1, \ b = 6, \ c = -1\\\\\Delta = b^{2}-4ac = 6^{2}-4\cdot(-1)\cdot(-1) = 36-4 = 32\\\\W = (p, q)\\\\p = \frac{-b}{2a} = \frac{-6}{-2} = 3, \ \ \ q = \frac{-\Delta}{4a} = \frac{-32}{-4} = 8\\\\W = (3, 8)[/tex]
[tex]c)\\y = 2x^{2}+4x-5\\\\a = 2, \ b = 4, \ c = -5\\\\\Delta = b^{2}-4ac = 4^{2}-4\cdot2\cdot(-5) = 16 + 40 = 56\\\\W = (p,q)\\\\p = \frac{-b}{2a} = \frac{-4}{4} = -1, \ \ \ q = \frac{-\Delta}{4a} = \frac{-56}{8} = -7\\\\W = (-1,-7)[/tex]
[tex]d)\\y = -3x^{2}+6x-2\\\\a = -3, \ b = 6, \ c = -2\\\\\Delta = b^{2}-4ac = 6^{2}-4\cdot(-3)\cdot(-2) = 36 - 24 = 12\\\\W = (p, q)\\\\p = \frac{-b}{2a} = \frac{-6}{-6} = 1, \ \ \ q = \frac{-\Delta}{4a} = \frac{-12}{-12} = 1\\\\W = (1,1)[/tex]
1.
[tex]a)\\y = -3x^{2}+6x-4 \\\\y= -3\cdot2^{2}+6\cdot2 - 4 = -3\cdot4+12-4 = -12+12-4 = -4\\\\y = -3\cdot(-1)^{2}+6\cdot(-1)-4 = -3\cdot1 -6-4 = -3-10 = -13[/tex]
[tex]a)\\y = 2x^{2}-3x-5\\\\y = 2\cdot2^{2}-3\cdot2 - 5 = 2\cdot4-6-5 = 8-11 = -3\\\\y = 2\cdot(-1)^{2}-3\cdot(-1)-5 = 2\cdot1 + 3 - 5 = 2-2 = 0[/tex]