Question

determine as coordenadas do vértice da parábola que representa as funções quadráticas a seguir: a) Y=X2-2X-3 b) Y=-X2+3X-5 c) Y=X2-4X+3

Answer 1

Vamos lá Luana:

$x_{v}=\frac{-b}{2a}\\ \\ y_{v}=\frac{-\Delta}{4a}\\ \\ a)\\ \\ y=x^{2}-2x-3\\ \\ \Delta=(-2)^{2}-4\cdot 1\cdot (-3)\\ \\ \Delta=4+12\\ \\ \Delta=16\\ \\ x_{v}=\frac{-(-2)}{2\cdot 1}\to x_{v}=\frac{2}{2}\to \boxed{x_{v}=1}\\ \\ y_{v}=\frac{-16}{4\cdot 1}\to y_{v}=\frac{-16}{4}\to \boxed{y_{v}=-4}\\ \\ \text{coordenadas do v\' ertice:}\\ \\ \boxed{V(1,-4)}\\ \\ b)\\ \\ y=-x^{2}+3x-5\\ \\ \Delta=3^{2}-4\cdot (-1)\cdot (-5)\\ \\ \Delta=9-20\\ \\ \Delta=-11$

$x_{v}=\frac{-3}{2\cdot (-1)}\to \boxed{x_{v}=\frac{3}{2}}\\ \\ y_{v}=\frac{-(-11)}{4\cdot (-1)}\to \boxed{y_{v}=-\frac{11}{4}}\\ \\ \text{coordenadas do v\' ertice:}\\ \\ \boxed{V(\frac{3}{2},-\frac{11}{4})}\\ \\ c)\\ \\ y=x^{2}-4x+3\\ \\ \Delta=(-4)^{2}-4\cdot 1\cdot 3\\ \\ \Delta=16-12\\ \\ \Delta=4\\ \\ x_{v}=\frac{-(-4)}{2\cdot 1}\to x_{v}=\frac{4}{2}\to \boxed{x_{v}=2}\\ \\ y_{v}=\frac{-4}{4\cdot 1}\to y_{v}=\frac{-4}{4}\to \boxed{y_{v}=-1}\\ \\ \text{coordenadas do v\' ertice:}$

$\boxed{V(2,-1)}$

Espero ter ajudado.