Question

1- Determine as coordenadas do vértice:
a) x²-4x-5
b) x²+2x-8
c) x²+4x
d )x²+2x+3
e) x²-2x+1

Answer 1

Resolução da questão, veja como se procede:

Item ''A'':

$\textsf{Coordenadas~do~v\'ertice~da~equa\c{c}\~ao}~~x^2-4x-5\\\\\\\ Xv = \dfrac{-b}{2a}\\\\\\ Xv = \dfrac{4}{2~\cdot~1}}\\\\\ Xv = \dfrac{-b}{2a}\\\\\\ Xv = \dfrac{4}{2}\\\\\\ \large\boxed{\boxed{\boxed{Xv = 2.}}}}\\\\\\\ \Delta = b^2-4~\cdot~a~\cdot~c\\\\ \Delta = (-4)^2-4~\cdot~1~\cdot~(-5)\\\\\ \Delta =16 + 20\\\\\ \Delta =36.\\\\\\\ Yv = \dfrac{-\Delta}{4a}\\\\\\ Yv = \dfrac{-36}{4~\cdot~1}}\\\\\\ Yv = \dfrac{-36}{4}}\\\\\\ \large\boxed{\boxed{\boxed{Yv = -9.}}}}}}}}$


Item ''B'':

$\textsf{Coordenadas~do~v\'ertice~da~equa\c{c}\~ao}~~x^2+2x-8\\\\\\\ Xv = \dfrac{-b}{2a}\\\\\\ Xv = \dfrac{-2}{2~\cdot~1}}\\\\\\ Xv = \dfrac{-2}{2}}\\\\\\\ \large\boxed{\boxed{\boxed{Xv = -1.}}}}}\\\\\\\\ \Delta = b^2-4~\cdot~a~\cdot~c\\\\ \Delta = 2^2-4~\cdot~1~\cdot~(-8)\\\\\ \Delta = 4 +32\\\\\ \Delta = 36\\\\\\\ Yv = \dfrac{-\Delta}{4a}\\\\\\ Yv = \dfrac{-36}{4~\cdot~1}}\\\\\\ Yv = \dfrac{-36}{4}}\\\\\\ \large\boxed{\boxed{\boxed{Yv = -9.}}}}}}}}$

Item ''C'':

$\textsf{Coordenadas~do~v\'ertice~da~equa\c{c}\~ao}~~x^2+4x\\\\\\\ Xv = \dfrac{-b}{2a}\\\\\\ Xv = \dfrac{-4}{2~\cdot~1}}\\\\\\ Xv = \dfrac{-4}{2}}\\\\\\\ \large\boxed{\boxed{\boxed{Xv = -2.}}}}\\\\\\\\ \Delta = b^2-4~\cdot~a~\cdot~c\\\\ \Delta = 4^2-4~\cdot~1~\cdot~0\\\\\ \Delta =16-0\\\\\ \Delta =16\\\\\\\ Yv = \dfrac{-\Delta}{4a}\\\\\\ Yv = \dfrac{-16}{4~\cdot~1}}\\\\\\ Yv = \dfrac{-16}{4}}\\\\\\ \large\boxed{\boxed{\boxed{Yv = -4.}}}}}$

Item ''D'':

$\textsf{Coordenadas~do~v\'ertice~da~equa\c{c}\~ao}~~x^2+2x+3\\\\\\\ Xv = \dfrac{-b}{2a}\\\\\\ Xv = \dfrac{-2}{2~\cdot~1}}\\\\\\ Xv = \dfrac{-2}{2}}\\\\\\\ \large\boxed{\boxed{\boxed{Xv = -1.}}}}}}\\\\\\\\ \Delta = b^2-4~\cdot~a~\cdot~c\\\\ \Delta = 2^2-4~\cdot~1~\cdot~3\\\\\ \Delta = 4-12\\\\\ \Delta = -8\\\\\\\ Yv = \dfrac{-\Delta}{4a}\\\\\\ Yv = \dfrac{8}{4~\cdot~1}}\\\\\\ Yv = \dfrac{8}{4}}\\\\\\\ \large\boxed{\boxed{\boxed{Yv = 2.}}}$

Item ''E'':

$\textsf{Coordenadas~do~v\'ertice~da~equa\c{c}\~ao}~~x^2-2x+1\\\\\\\ Xv = \dfrac{-b}{2a}\\\\\\ Xv = \dfrac{2}{2~\cdot~1}}\\\\\\ Xv = \dfrac{2}{2}}\\\\\\\ \large\boxed{\boxed{\boxed{Xv = 1.}}}}\\\\\\\\ \Delta = b^2-4~\cdot~a~\cdot~c\\\\ \Delta =(-2)^2-4~\cdot~1~\cdot~1\\\\\ \Delta = 4-4\\\\\ \Delta =0~\to~\textsf{Se~Delta~for~igual~a~Zero~Xv~\'e~igual~a~Yv,~ou~seja}:\\\\\\ \large\boxed{\boxed{\boxed{Xv=Yv = 1.}}}}}}$

Espero que te ajude. :-).

Dúvidas? Comente.