45 Вправа 45 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 321

45. Побудуйте графік функції:

1) ${\displaystyle y=}$${\displaystyle \frac{x^2-3x+2}{x-2}}$;

ОДЗ: ${\displaystyle x-2\ne 0}$;

${\displaystyle x\ne 2}$;

${\displaystyle x^2-3x+2=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle 3}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle 2}$;

${\displaystyle x_1=}$${\displaystyle 1}$; ${\displaystyle x_2=}$${\displaystyle 2}$;

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

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

${\displaystyle y=}$${\displaystyle x-1}$:

Графіком є пряма ${\displaystyle y=}$${\displaystyle x-1}$ з «виколотою» точкою ${\displaystyle (2;1)}$:

2) ${\displaystyle y=}$${\displaystyle \frac{3x^2+2x-1}{x+1}}$;

ОДЗ: ${\displaystyle x+1\ne 0}$;

${\displaystyle x\ne -1}$;

${\displaystyle 3x^2+2x-1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 2^2-4\cdot 3\cdot (-1)=}$${\displaystyle 4+12=}$${\displaystyle 16}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-2\pm \sqrt{16}}{2\cdot 3}=}$${\displaystyle \frac{-2\pm 4}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-2+4}{6}=}$${\displaystyle \frac{2}{6}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-2-4}{6}=}$${\displaystyle -1}$;

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

${\displaystyle y=}$${\displaystyle \frac{\cancel{\left(x+1\right)}(3x-1)}{\cancel{\left(x+1\right)}}=}$${\displaystyle 3x-1}$, де ${\displaystyle x\ne -1}$;

${\displaystyle y=}$${\displaystyle 3x-1}$:

Графіком є пряма ${\displaystyle y=}$${\displaystyle 3x-1}$ з «виколотою» точкою ${\displaystyle (-1;-4)}$: