Question

determine o vertice f(x) = x2 + 6 x +5​

Answer 1

As coordenadas do vértice da função do 2° grau são dadas por:

$(V_x~,~V_y)~=~\left(-\dfrac{b}{2a}~,~-\dfrac{\Delta}{4a}\right)$

$Onde~~~\Delta=b^2-4ac$

Os coeficientes da função dada são:

$\rightarrow~~a=1\\\\\rightarrow~~b=6\\\\\rightarrow~~c=5$

Substituindo os coeficientes, podemos achar as coordenadas do vértice:

$(V_x~,~V_y)~=~\left(-\dfrac{6}{2~.~1}~,~-\dfrac{6^2-4.1.5}{4~.~1}\right)\\\\\\\\(V_x~,~V_y)~=~\left(-\dfrac{6}{2}~,~-\dfrac{36-20}{4}\right)\\\\\\\\(V_x~,~V_y)~=~\left(-3~,~-\dfrac{16}{4}\right)\\\\\\\\\boxed{(V_x~,~V_y)~=~\left(-3~,~-4\right)}$

Answer 2

Equacão = f(x) = x² + 6 x +5

a = 1

b = 6

c = 5

= V = (p,q)

p = - B/2.A q = - ∆/4.A

p = - 6/2.1 q = - 16/4.1

p = -3 q = - 4

V = (-3, -4)