Question

gráfico da equação x2-3x-4=0 com vertices

Answer 1

$\mathrm{f(x)=x^2-3x-4\ \to\ a=1\ \| \ b=-3\ \| \ c=-4}\\\\ \textrm{Encontrando as ra\'izes de f(x):}\\\\ \mathrm{0=x^2-3x-4}\\\\ \textrm{Utilizando a f\'ormula quadr\'atica, obteremos:}\\\\ \mathrm{x=\cfrac{-b\pm\sqrt{\Delta}}{2a}\ \to\ \Delta=b^2-4ac}\\\\ \mathrm{\Delta=(-3)^2-4.1.(-4)=9+16=25}\\\\ \mathrm{x_1=\cfrac{-(-3)+\sqrt{25}}{2.1}=\cfrac{3+5}{2}=\cfrac{8}{2}=4}\\\\ \mathrm{x_2=\cfrac{-(-3)-\sqrt{25}}{2.1}=\cfrac{3-5}{2}=\cfrac{-2}{2}=-1}\\\\ \mathrm{Ra\'izes\ \to\ S=\{4,-1\}}}$

$\textrm{Encontrando o v\'ertice de f(x):}\\\\ \mathrm{x_v=\cfrac{-b}{2a}=\cfrac{-(-3)}{2.1}=\cfrac{3}{2}}\\\\ \mathrm{y_v=\cfrac{-\Delta}{4a}=\cfrac{-25}{4.1}}=\cfrac{-25}{4}}\\\\ \mathrm{V\'ertice\ \to\ v=\bigg(\cfrac{3}{2},\cfrac{-25}{4}\bigg)}\\\\ \textrm{Ponto de intersec\c{c}\~ao no eixo das ordenadas:}\\\\ \mathrm{x=0\ \to\ f(x)=0^2-3.0-4\ \to\ f(x)=-4}\\\\ \textrm{Intersec\c{c}\~ao\ em\ y}\ \to\ \mathrm{(0,-4)}}$