Вправи: 400

400. Для функції y = x² + 6x знайдіть: 1) значення функції, якщо x = 0; –1; 3; –4; 2) значення аргументу, якщо y = 0; 7; 16.

1) x = 0: 0² + 6 • 0 = y; y = 0.

x = –1: y = (–1)² + 6(–1); y = 1 – 6; y = –5.

x = 3: y = 3² + 6 • 3; y = 9 + 18; y = 27.

x = –4: y = (–4)² + 6(–4); y = 16 – 24; y = –8.

2) y = 0: x² + 6x = 0;

x(x + 6) = 0; x = 0;

x + 6 = 0; x = –6.

y = 7: x² + 6x = 7;

x² + 6x – 7 = 0; D = b² – 4ac;

D = 36 – 4 • (–7) = 36 + 28 = 64;

x_1,2 = -b ± D2a; x₁ = -6+82 = 1; x₂ = -6-82 = –7.

y = 16: x² + 6x = 16;

x² + 6x – 16 = 0; D = b² – 4ac;

D = 36 – 4 • (–16) = 36 + 64 = 100;

x_1,2 = -6 ± 102; x₁ = -6+102 = 42 = 2; x₂ = -6-102 = -162 = –8.