Question

Determine as coordenadas do vértice , das parábolas das funções.

a- f (x) = 8² - X² + -5

b- f (x) = x²

c- f (x) = x² - 4x +3

Answer 1

$\mathsf{Equa\c{c}\~ao~quadr\'atica:~\fbox{ \mathsf{ax^2+bx+c} }}\\\\\\\mathsf{Coordenadas~do~v\'ertice~da~par\'abola:}\\\\\\\mathsf{X_v= \dfrac{-b}{2a}}\\\\\\\mathsf{Yv=\dfrac{-\Delta}{4a}~~~onde~~~\Delta=b^2-4ac}\\\\\\\textsf{Coordenadas da primeira par\'abola dada por:~~}\mathsf{f(x)=-x^2+8^2-5}\\\\\\\mathsf{X_v=0~~(visto~que~b=0)}\\\\\\\mathsf{Y_v=\dfrac{-(0^2-4\cdot (-1)\cdot(59)}{4\cdot(-1)}~\Rightarrow~Y_v=\dfrac{-236}{-4}~\Rightarrow~Y_v=59}\\\\\\\mathsf{Coordenadas:~(0;59)}$

$\mathsf{Coordenadas~do~v\'ertice~da~par\'abola~dada~por:~f(x)=x^2}\\\\\\\mathsf{X_v=0~~(visto~que~b=0)}\\\\\\\mathsf{Y_v=0~(visto~que~b=0~e~c=0)}\\\\\\\mathsf{Coordenadas~(0;0)}$


$\mathsf{Coordenadas~do~v\'ertice~da~par\'abola~dada~por:~x^2-4x+3}\\\\\\\mathsf{X_v=\dfrac{-(-4)}{2\cdot 1}~\Rightarrow~X_v=\dfrac{4}{2}=2}\\\\\\\mathsf{Y_v=\dfrac{-((-4)^2-4\cdot 1\cdot 3)}{4\cdot 1}~\Rightarrow~Y_v=\dfrac{-(16-12)}{4}~\Rightarrow~Y_v=-\dfrac{4}{4}=-1}\\\\\\\mathsf{Coordenadas:~(2;-1)}$


$\textsf{Os respectivos gr\'aficos est\~ao em anexo:}$