1093 Вправа 1093Алгебра 8 клас Бевз 2025 · § 21 · стор. 205

1093. Побудуй графік функції. а) y = (x² − x − 12)/(x − 4); б) y = (3x² − 7x + 2)/(2 − x).

а) ОВ: ${\displaystyle x\ne 4}$;

${\displaystyle x^2-x-12}$: ${\displaystyle x_1+x_2=}$${\displaystyle 1}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -12}$ → ${\displaystyle x_1=}$${\displaystyle 4,x_2=}$${\displaystyle -3}$;

${\displaystyle x^2-x-12=}$${\displaystyle \left(x-4\right)\left(x+3\right)}$;

${\displaystyle y=}$${\displaystyle \frac{\cancel{\left(x-4\right)}(x+3)}{\cancel{x-4}}=}$${\displaystyle x+3}$;

Графік — пряма ${\displaystyle y=}$${\displaystyle x+3}$ без точки ${\displaystyle (4;7)}$:

б) ОВ: ${\displaystyle x\ne 2}$;

${\displaystyle 3x^2-7x+2}$: ${\displaystyle D=}$${\displaystyle {(-7)}^2-4\cdot 3\cdot 2=}$${\displaystyle 25}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{7\pm 5}{6}}$;

${\displaystyle x_1=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle 3x^2-7x+2=}$${\displaystyle 3(x-2)(x-\frac{1}{3})=}$${\displaystyle \left(x-2\right)\left(3x-1\right)}$;

${\displaystyle y=}$${\displaystyle \frac{\cancel{\left(x-2\right)}(3x-1)}{-\cancel{\left(x-2\right)}}=}$${\displaystyle 1-3x}$;

Графік — пряма ${\displaystyle y=}$${\displaystyle 1-3x}$ без точки ${\displaystyle (2;-5)}$: