992 Вправа 992Алгебра 8 клас Бевз 2025 · § 19 · стор. 186

992. Не виконуючи побудови, знайди координати точок перетину графіків функцій. а) y = 2x² + 7x − 2 і y = x² − 2x − 16; б) y = 6/x і y = x − 5.

Прирівнюємо праві частини, знаходимо ${\displaystyle x}$, потім ${\displaystyle y}$.

а) ${\displaystyle 2x^2+7x-2=}$${\displaystyle x^2-2x-16}$;

${\displaystyle 2x^2+7x-2-x^2+2x+16=}$${\displaystyle 0}$;

${\displaystyle x^2+9x+14=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 9,c=}$${\displaystyle 14}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 9^2-4\cdot 1\cdot 14=}$${\displaystyle 81-56=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-9\pm \sqrt{25}}{2}=}$${\displaystyle \frac{-9\pm 5}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-9+5}{2}=}$${\displaystyle -2}$;

${\displaystyle x_2=}$${\displaystyle \frac{-9-5}{2}=}$${\displaystyle -7}$;

${\displaystyle y(-2)=}$${\displaystyle {(-2)}^2-2\cdot (-2)-16=}$${\displaystyle 4+4-16=}$${\displaystyle -8}$;

${\displaystyle y(-7)=}$${\displaystyle {(-7)}^2-2\cdot (-7)-16=}$${\displaystyle 49+14-16=}$${\displaystyle 47}$;

Точки: ${\displaystyle \left(-2;-8\right)}$, ${\displaystyle \left(-7;47\right)}$;

б) ${\displaystyle \frac{6}{x}=}$${\displaystyle x-5}$, ${\displaystyle x\ne 0}$;

${\displaystyle 6=}$${\displaystyle x^2-5x}$;

${\displaystyle 0=}$${\displaystyle x^2-5x-6}$;

${\displaystyle x^2-5x-6=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -5,c=}$${\displaystyle -6}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-5)}^2-4\cdot 1\cdot (-6)=}$${\displaystyle 25+24=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{5\pm \sqrt{49}}{2}=}$${\displaystyle \frac{5\pm 7}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+7}{2}=}$${\displaystyle 6}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-7}{2}=}$${\displaystyle -1}$;

${\displaystyle y\left(6\right)=}$${\displaystyle 6-5=}$${\displaystyle 1}$;

${\displaystyle y(-1)=}$${\displaystyle -1-5=}$${\displaystyle -6}$;

Точки: ${\displaystyle \left(6;1\right)}$, ${\displaystyle \left(-1;-6\right)}$.