Question

0=x²-4x-3 qual as coordenadas do vertice dessa parab

Answer 1

$f(x)=x^2-4x-3$

$x_V=\dfrac{-b}{2a}$

$x_V=\dfrac{-(-4)}{2\cdot1}=\dfrac{4}{2} \iff \boxed{x_V=2}$

$y_V=\dfrac{-\Delta}{4a}$

$\Delta=(-4)^2-4\cdot1\cdot(-3)=16+12=28$

$y_V=\dfrac{-28}{4} \iff \boxed{y_V=-7}$

$V(2,-7)$

Answer 2

Bom dia, usarei a ideia de derivadas para achar o vértice..

f(x) = x² - 4x - 3 

f'(x) = 2x - 4

f'(x) = 0 

2x - 4 = 0 

2x = 4 

x = 4 / 2

x = 2

substitui x do vértice na função para achar y

f(2) = 2² - 4.2 - 3

f(2) = 4 - 8 - 3

f(2) = 4 - 11

f(2) = - 7 

coordenada do vértice (2, - 7)