Question

a) Determine as raízes (zeros) da função de 2o grau x2 - 5x + 6 = 0


b) Determine a raiz (zero) da função de 1o grau
f(x) = x/2 – 4


c) Seja a função f(x) = x2 – 5x + 6 = 0 , determine Δ e as coordenadas do vértice: Sabemos que:
Δ = b2 - 4.a.c xV = -b / 2.a yV = - Δ / 4.a

Answer 1

a)

$\sf x^2-5x+6=0$

$\sf \Delta=(-5)^2-4\cdot1\cdot6$

$\sf \Delta=25-24$

$\sf \Delta=1$

$\sf y=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}=\dfrac{5\pm1}{2}$

$\sf x'=\dfrac{5+1}{2}~\rightarrow~x'=\dfrac{6}{2}~\rightarrow~x'=3$

$\sf x"=\dfrac{5-1}{2}~\rightarrow~x"=\dfrac{4}{2}~\rightarrow~x"=2$

As raízes dessa função são $\sf 2~e~3$

b)

$\sf \dfrac{x}{2}-4=0$

$\sf \dfrac{x}{2}=4$

$\sf x=2\cdot4$

$\sf x=8$

A raiz dessa função é $\sf 8$

c)

• $\sf \Delta=b^2-4\cdot a\cdot c$

$\sf \Delta=(-5)^2-4\cdot1\cdot6$

$\sf \Delta=25-24$

$\sf \Delta=1$

• $\sf x_V=\dfrac{-b}{2a}$

$\sf x_V=\dfrac{-(-5)}{2\cdot1}$

$\sf x_V=\dfrac{5}{2}$

• $\sf y_V=\dfrac{-\Delta}{4a}$

$\sf y_V=\dfrac{-1}{4\cdot1}$

$\sf y_V=\dfrac{-1}{4}$