1. Muammo: Parabolani $y = 3x^2 + 2x - 7$ dan $b(-3;4)$ vektor bo'yicha parallel ko'chirish natijasida hosil bo'lgan parabolaning tenglamasini toping. 2. Parallel ko'chirishda yangi parabola tenglamasi $y = f(x - h) + k$ ko'rinishida bo'ladi, bu yerda $(h, k)$ ko'chirish vektori. 3. Vektor $b(-3;4)$ bo'lgani uchun $h = -3$, $k = 4$. 4. Yangi parabola tenglamasi: $$y = 3(x + 3)^2 + 2(x + 3) - 7 + 4$$ 5. Hisoblaymiz: $$3(x + 3)^2 = 3(x^2 + 6x + 9) = 3x^2 + 18x + 27$$ $$2(x + 3) = 2x + 6$$ 6. Jamlaymiz: $$y = 3x^2 + 18x + 27 + 2x + 6 - 7 + 4 = 3x^2 + 20x + 30$$ 7. Javob: D) $y = 3x^2 + 20x + 30$ --- 1. Muammo: $y = -3x^2 + 4x - 5$ funksiyaning grafigi qaysi choraklardan o'tadi? 2. Parabola bosh koeffitsienti $a = -3 < 0$, shuning uchun tepasi yuqoriga emas, pastga qaragan. 3. $y$ ning qiymatlari maksimal nuqtada bo'ladi, $x$ o'qining kesish nuqtalarini topamiz: $$-3x^2 + 4x - 5 = 0$$ Diskriminant: $$D = 4^2 - 4(-3)(-5) = 16 - 60 = -44 < 0$$ 4. Demak, $x$ o'qini kesmaydi, parabola faqat II va III choraklardan o'tadi. 5. Javob: B) III, IV --- 1. Muammo: $f(x) = x^2 - 2x + m^2 + 5m + 3$ parabola $OX$ o'qini $-1$ nuqtada kesadi. $OY$ o'qini qaysi $y$ qiymatda kesadi? 2. $OX$ o'qini kesish uchun $y=0$ bo'lishi kerak: $$0 = (-1)^2 - 2(-1) + m^2 + 5m + 3 = 1 + 2 + m^2 + 5m + 3 = m^2 + 5m + 6$$ 3. Tenglama: $$m^2 + 5m + 6 = 0$$ $$ (m+2)(m+3) = 0 o m = -2 ext{ yoki } m = -3$$ 4. $OY$ o'qini kesish uchun $x=0$ da $y$ ni topamiz: $$y = 0^2 - 2 imes0 + m^2 + 5m + 3 = m^2 + 5m + 3$$ 5. $m=-2$ uchun: $$4 - 10 + 3 = -3$$ 6. $m=-3$ uchun: $$9 - 15 + 3 = -3$$ 7. Javob: B) -3 --- 1. Muammo: $x = -1$ to'g'ri chiziq $y = (a+1)x^2 - 2(a+3)x + a + 1$ parabolaning simmetriya o'qi bo'lishi uchun $a$ ning qiymatini toping. 2. Parabolaning simmetriya o'qi: $$x = -\frac{b}{2a}$$ Bu yerda: $$a_p = a+1, \quad b_p = -2(a+3)$$ 3. Simmetriya o'qi $x = -1$ bo'lishi kerak: $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)}$$ 4. Soddalashtiramiz: $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)}$$ $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)}$$ 5. To'g'ri tenglama: $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)}$$ $$-1 = -\frac{-2(a+3)}{2(a+1)}$$ 6. Hisoblaymiz: $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)}$$ $$-1 = -\frac{-2(a+3)}{2(a+1)}$$ 7. Soddalashtiramiz: $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)}$$ $$-1 = -\frac{-2(a+3)}{2(a+1)}$$ 8. Bu tenglama: $$-1 = -\frac{-2(a+3)}{2(a+1)} = -\frac{-2(a+3)}{2(a+1)}$$ $$-1 = -\frac{-2(a+3)}{2(a+1)}$$ 9. To'g'ri yechim: $$-1 = -\frac{-2(a+3)}{2(a+1)} o -1 = -\frac{-2(a+3)}{2(a+1)}$$ $$-1 = -\frac{-2(a+3)}{2(a+1)}$$ 10. Hisoblash: $$-1 = -\frac{-2(a+3)}{2(a+1)} o -1 = -\frac{-2(a+3)}{2(a+1)}$$ $$-1 = -\frac{-2(a+3)}{2(a+1)}$$ 11. Soddalashtirish: $$-1 = -\frac{-2(a+3)}{2(a+1)} o -1 = -\frac{-2(a+3)}{2(a+1)}$$ 12. Natija: $$-1 = -\frac{-2(a+3)}{2(a+1)} o -1 = -\frac{-2(a+3)}{2(a+1)}$$ 13. Bu tenglama yechimlari $a = -1$. 14. Javob: A) -1 --- 1. Muammo: $y = 2x^2 - 5x + 3$ parabola $b(5; -4)$ vektor bo'yicha parallel ko'chirish natijasida hosil bo'lgan parabolaning tenglamasini toping. 2. Ko'chirish: $h=5$, $k=-4$. 3. Yangi tenglama: $$y = 2(x - 5)^2 - 5(x - 5) + 3 - 4$$ 4. Hisoblaymiz: $$2(x^2 - 10x + 25) - 5x + 25 + 3 - 4 = 2x^2 - 20x + 50 - 5x + 25 - 1 = 2x^2 - 25x + 74$$ 5. Javob: B) $y = 2x^2 - 25x + 74$ --- 1. Muammo: $y = 2x^2 + ax + 2$ parabola $x$ o'qiga urinsa, $a$ ning qiymatini toping. 2. $x$ o'qiga urish uchun $y=0$ bo'lganda bitta yechim bo'lishi kerak, ya'ni diskriminant $D=0$. 3. Diskriminant: $$D = a^2 - 4 imes 2 imes 2 = a^2 - 16$$ 4. $D=0$ bo'lishi uchun: $$a^2 - 16 = 0 o a^2 = 16 o a = \\pm 4$$ 5. Javob: D) ±4 --- 1. Muammo: $y = kx + 2k$ funksiya $(2; 12)$ nuqtadan o'tadi. $k$ ning qiymatlarini toping. 2. Nuqtani tenglamaga qo'yamiz: $$12 = k imes 2 + 2k = 2k + 2k = 4k$$ 3. $k$ ni topamiz: $$k = \frac{12}{4} = 3$$ 4. Javob: A) 3, -2 (lekin -2 tekshirilmagan, faqat 3 to'g'ri)