Odpowiedź:
15. f(x) = -5[tex]x^{2}[/tex] + x + 4
Δ = 1 - 4(-5)(4) = 1 + 80 = 81
[tex]\sqrt{delta}[/tex] = 9
x1 = (-1 - 9)/-10 = -10/-10 = 1
x2 = (-1 + 9)/-10 = 8/-10 = -4/5
p = -1/-10 = 1/10
q = -9/-20 = 9/20
Kanoniczna = -5(x - 1/10)^2 + 9/20
Iloczynowa = -5(x - 1)(x + 4/5)
16. f(x) = [tex]x^{2}[/tex] - 8x + 15
Δ = 64 - 4(15) = 4
[tex]\sqrt{delta}[/tex] = 2
x1 = (8 - 2)/2 = 6/2 = 3
x2 = (8 + 2)/2 = 10/2 = 5
p = 8/2 = 4
q = -2/4 = -1/2
Kanoniczna = (x - 4)^2 - 1/2
Iloczynowa = (x - 3)(x - 5)
17. f(x) = -[tex]x^{2}[/tex] + 7x - 12
Δ = 49 - 4(-1)(-12) = 49 - 48 = 1
[tex]\sqrt{delta}[/tex] = 1
x1 = (-7 - 1)/-2 = -8/-2 = 4
x2 = (-7 + 1)/-2 = -6/-2 = 3
p = -7/-2 = 7/2
q = -1/-4 = 1/4
Kanoniczna = -(x - 7/2)^2 + 1/4
Iloczynowa = -(x - 4)(x - 3)
18. f(x) = [tex]x^{2}[/tex] - 3x - 4
Δ = 9 - 4(-4) = 9 + 16 = 25
[tex]\sqrt{delta}[/tex] = 5
x1 = (3 - 5)/2 = -2/2 = -1
x2 = (3 + 5)/2 = 8/2 = 4
p = 3/2
q = -5/4
Kanoniczna = (x - 3/2)^2 - 5/4
Iloczynowa = (x + 1)(x - 4)
19. f(x) = -3[tex]x^{2}[/tex] + 2x + 5
Δ = 4 - 4(-3)(5) = 4 + 60 = 64
[tex]\sqrt{delta}[/tex] = 8
x1 = (-2 - 8)/-6 = -10/-6 = 5/3
x2 = (-2 + 8)/-6 = 6/-6 = -1
p = -2/-6 = 1/3
q = -8/-12 = 2/3
Kanoniczna = -3(x - 1/3)^2 + 2/3
Iloczynowa = -3(x - 5/3)(x + 1)
