976 Вправа 976Алгебра 8 клас Бевз 2025 · § 19 · стор. 184

976. Не виконуючи побудови, знайди координати точок перетину графіка функції з осями координат. а) y = x² + x − 20; б) y = 0,5x² − 5x + 8.

З віссю ${\displaystyle Oy}$: ${\displaystyle x=}$${\displaystyle 0}$; з віссю ${\displaystyle Ox}$: ${\displaystyle y=}$${\displaystyle 0}$.

а) ${\displaystyle y=}$${\displaystyle x^2+x-20}$;

${\displaystyle Oy}$: ${\displaystyle y=}$${\displaystyle 0^2+0-20=}$${\displaystyle -20}$ → ${\displaystyle \left(0;-20\right)}$;

${\displaystyle Ox}$:

${\displaystyle x^2+x-20=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 1,c=}$${\displaystyle -20}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 1^2-4\cdot 1\cdot (-20)=}$${\displaystyle 1+80=}$${\displaystyle 81}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{81}}{2}=}$${\displaystyle \frac{-1\pm 9}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+9}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-9}{2}=}$${\displaystyle -5}$;

→ ${\displaystyle \left(4;0\right)}$, ${\displaystyle \left(-5;0\right)}$;

б) ${\displaystyle y=}$${\displaystyle 0,5x^2-5x+8}$;

${\displaystyle Oy}$: ${\displaystyle y=}$${\displaystyle 8}$ → ${\displaystyle \left(0;8\right)}$;

${\displaystyle Ox}$:

${\displaystyle 0,5x^2-5x+8=}$${\displaystyle 0\mid \cdot 2}$;

${\displaystyle x^2-10x+16=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -10,c=}$${\displaystyle 16}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-10)}^2-4\cdot 1\cdot 16=}$${\displaystyle 100-64=}$${\displaystyle 36}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{10\pm \sqrt{36}}{2}=}$${\displaystyle \frac{10\pm 6}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{10+6}{2}=}$${\displaystyle 8}$;

${\displaystyle x_2=}$${\displaystyle \frac{10-6}{2}=}$${\displaystyle 2}$;

→ ${\displaystyle \left(8;0\right)}$, ${\displaystyle \left(2;0\right)}$.