Question

me ajudem!!!!
veja o anexo explicando .

Exercícios:
Faça o estudo dos sinais das funções quadráticas abaixo:
a) x2-4x +3

b) b) x2-5x+6

9a0ca4e633d13f324d6b1d9d9592a551.pdf

Answer 1

Explicação passo-a-passo:

a) $\sf y=x^2-4x+3$

$\sf x^2-4x+3=0$

$\sf \Delta=(-4)^2-4\cdot1\cdot3$

$\sf \Delta=16-12$

$\sf \Delta=4$

$\sf x=\dfrac{-(-4)\pm\sqrt{4}}{2\cdot1}=\dfrac{4\pm2}{2}$

$\sf x'=\dfrac{4+2}{2}~\Rightarrow~x'=\dfrac{6}{2}~\Rightarrow~\red{x'=3}$

$\sf x"=\dfrac{4-2}{2}~\Rightarrow~x"=\dfrac{2}{2}~\Rightarrow~\red{x"=1}$

Temos que:

=> $\sf y > 0,~para~x < 1~ou~x > 3$

=> $\sf y < 0,~para~1 < x < 3$

=> $\sf y=0,~para~x=1~ou~x=3$

b) $\sf y=x^2-5x+6$

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

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

$\sf \Delta=25-24$

$\sf \Delta=1$

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

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

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

Temos que:

=> $\sf y > 0,~para~x < 2~ou~x > 3$

=> $\sf y < 0,~para~2 < x < 3$

=> $\sf y=0,~para~x=2~ou~x=3$

Answer 2

a) x² - 4x + 3 = 0

a = 1, b = - 4, c = 3

∆ = b² - 4ac

∆ = ( - 4 )² - 4 • 1 • 3

∆ = 16 - 12

∆ = 4

x = - b ± √∆ / 2a

x = - ( - 4 ) ± √4 / 2 • 1

x = 4 ± 2 / 2

x' = 4 + 2/2

x' = 6/2

x' = 3

x" = 4 - 2/2

x" = 2/2

x" = 1

S = { 1, 3 }

b) x² - 5x + 6 = 0

a = 1

b = -5

c = 6

Delta:

Δ = b² - 4ac

Δ = (-5)² - 4 * 1 * 6

Δ = 25 - 24

Δ = 1

Bhaskara:

x = - b ± √Δ / 2a

x = - (-5) ± √1 / 2 * 1

x = 5 ± 1 / 2

x' = 5 + 1 / 2 = 6 / 2 = 3

x'' = 5 - 1 / 2 = 4 / 2 = 2

S = {2, 3}

Espero ter ajudado ^_^