977 Вправа 977Алгебра 8 клас Бевз 2025 · § 19 · стор. 184

977. Не виконуючи побудови, знайди координати точок перетину графіка функції з осями координат. а) y = x² + 3x − 28; б) y = −2x² + 3x − 1.

а) ${\displaystyle y=}$${\displaystyle x^2+3x-28}$;

${\displaystyle Oy}$: ${\displaystyle y=}$${\displaystyle -28}$ → ${\displaystyle \left(0;-28\right)}$;

${\displaystyle Ox}$:

${\displaystyle x^2+3x-28=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 3,c=}$${\displaystyle -28}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 1\cdot (-28)=}$${\displaystyle 9+112=}$${\displaystyle 121}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-3\pm \sqrt{121}}{2}=}$${\displaystyle \frac{-3\pm 11}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3+11}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3-11}{2}=}$${\displaystyle -7}$;

→ ${\displaystyle \left(4;0\right)}$, ${\displaystyle \left(-7;0\right)}$;

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

${\displaystyle Oy}$: ${\displaystyle y=}$${\displaystyle -1}$ → ${\displaystyle \left(0;-1\right)}$;

${\displaystyle Ox}$:

${\displaystyle -2x^2+3x-1=}$${\displaystyle 0\mid \cdot \left(-1\right)}$;

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

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -3,c=}$${\displaystyle 1}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 2\cdot 1=}$${\displaystyle 9-8=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{1}}{4}=}$${\displaystyle \frac{3\pm 1}{4}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{3-1}{4}=}$${\displaystyle 0,5}$;

→ ${\displaystyle \left(1;0\right)}$, ${\displaystyle \left(0,5;0\right)}$.