Question

-x²-x+2?
função do segundo grau e os pontos no plano cartesiano.

Answer 1

f(x)=-x²-x+2

a=-1<0( admite valor máximo)

$\Delta={b}^{2}-4ac\\\Delta={(-1)}^{2}-4.(-1). 2=1+8=9$

coordenadas do vértice:

$\boxed{\boxed{\mathsf{x_{v}=-\dfrac{b}{2a}}}}$

$\boxed{\boxed{\mathsf{x_{v}=\dfrac{-[-1]}{2.(-1)}=-\dfrac{1}{2}}}}$

$\boxed{\boxed{\mathsf{y_{v}=-\dfrac{\Delta}{4a}}}}$

$\boxed{\boxed{\mathsf{y_{v}-\dfrac{9}{4(-1)}=\dfrac{9}{4}}}}$

$V(x_{v},y_{v})\rightarrow\,V(-\dfrac{1}{2},\dfrac{9}{4})$

raízes:

$x=\dfrac{-b\pm\sqrt{\Delta}}{2a} \\x=\dfrac{1\pm3}{-2}$

$\boxed{\boxed{\mathsf{x'=\dfrac{1+3}{-2}=-2}}}$

$\boxed{\boxed{\mathsf{x''=\dfrac{1-3}{-2}=1}}}$

Intersecção com o eixo y:

$\boxed{\boxed{\mathsf{f(0)={0}^{2}-0+2=2}}}$

O gráfico esta anexo.