Last year, my husband Mike and I decided to visit New Zealand. We wanted to tour the
country, but we both hate long car journeys. The travel agent suggested a 13-day coach trip. It was a
good price, so we booked it with our plane tickets. We made a good choice. The coach journeys passed
quickly and our driver told us about each place. We learned a lot from him. We flew from London to
Christchurch and had a free day there before the coach trip started. We were not at all tired, so we
walked round the city from morning till night. It had good museums, many restaurants and lovely
shops. The best place we visited on the trip was Queenstown. You can choose to do almost anything,
from sailing to climbing. We had three days there, but it was not enough. All the hotels were good. My
favourite one was the Puka Park Lodge. It was on a hill above a beach and there were trees
everywhere. We woke up and listened to the birds singing. Now, when we are eating breakfast at home
and we hear the noise of the traffic, we think of those beautiful mornings in New Zealand!
Дана функция у= х²- 2х - 3.
График её - парабола ветвями вверх.
Находим её вершину: хо = -в/2а = 2/(2*1) = 1.
уо = 1 - 2 - 3 = -4.
В точке (1; -4) находится минимум функции.
а) промежутки возрастания и убывания функции:
убывает х ∈ (-∞; 1),
возрастает х ∈ (1; +∞).
б) наименьшее значение функции: в точке (1; -4) находится минимум функции уmin = -4.
в) при каких значениях х у > 0.
Для этого надо найти точки пересечения графиком оси Ох
(при этом у = 0).
х²- 2х - 3 = 0.
Квадратное уравнение, решаем относительно x:
Ищем дискриминант:
D=(-2)^2-4*1*(-3)=4-4*(-3)=4-(-4*3)=4-(-12)=4+12=16;
Дискриминант больше 0, уравнение имеет 2 корня:
x_1=(√16-(-2))/(2*1)=(4-(-2))/2=(4+2)/2=6/2=3;
x_2=(-√16-(-2))/(2*1)=(-4-(-2))/2=(-4+2)/2=-2/2=-1.
Функция (то есть у) больше 0 при х ∈ (-∞; -1) ∪ (3; +∞)
sinn/12-sin5n/12=sin15°-sin75°=sin(-60°)
Объяснение: