Question

2)Encontre as raízes e os vertices de X e Y das funções abaixo:
a)f(x) = x² + 3x – 1.
b) f(x) = -2x² + 4x

Answer 1

Resposta

$a) S = \left\{\frac{-3}{2},\frac{-13}{4}\right\}$  $V = \left\{\frac{-3}{2};\frac{-13}{4} \left\}$

b) $S = \left\{ 0,2\left\}$   $V = \left\{0,2\lleft\}$

Raízes e Vértices da Função de 2º Grau

a)f(x) = x² + 3x – 1

  Delta:

 Δ = 3² - 4.1.(-1)

 Δ = 9 + 4

 Δ = 13

 Raízes:

 $x = \frac{-3\pm\sqrt{13} }{2.1} \therefore x\simeq\frac{-3\pm3,6}{2}$

 $x'=\frac{-3+3,6}{2}\therefore x'\simeq\frac{0,6}{2}\therefore x'\simeq 0,3$

 $x'' \simeq \frac{-3-3,6}{2}\therefore x'' \simeq \frac{-6,6}{2}\therefore x'' \simeq - 3,3$

 S ≅ {0,3; - 3,3}

Vértices

 $Vx = \frac{-b}{2a} \therefore Vx = \frac{-3}{2}$

 $Vy = \frac{-\triangle}{4.1}\therefore Vy = \frac{-13}{4}$

b) f(x) = -2x² + 4x

   Delta:

   Δ = (-4)² - 4. (-2) .0

   Δ = 16

   Raízes:

   $x = \frac{-4\pm\sqrt{16} }{2.(-2)}\therefore x = \frac{-4\pm4}{-4}$

   $x' = \frac{-4+4}{-4}\therefore x' = \frac{0}{-4}\therefore x' = 0$

   $x'' = \frac{-4-4}{-4} \therefore x'' = \frac{-8}{-4}\therefore x'' = 2$

   $S = \left\{ 0,2\left\}$

  Vértices:

   $Vx = \frac{-b}{2a}\therefore Vx = \frac{-4}{2(-2)}\therefore Vx = \frac{-4}{-4}\therefore Vx = 1$

   $Vy = \frac{-\triangle}{4(-2)}\therefore Vy = \frac{-16}{-8}\therefore Vy = 2$

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Answer 2

Resposta:

a)f(x) = x² + 3x – 1

x'=[-3+√(9+4)]/2=(-3+√13)/2

x''=[-3-√(9+4)]/2=(-3-√13)/2

Vértice(xv,yv)

xv=(x'+x'')/2 =[(-3+√13)/2 +(-3-√13)/2]/2 =(-6/2)/2= -3/2

yv=f(xv)=f(-3/2)=(-3/2)²+3*(-3/2)-1 = 9/4-9/2-1 =-9/4-1= -13/4

b) f(x) = -2x² + 4x

-2x*(x-2)=0

-2x=0 ==>x'=0

x-2=0 ==>x''=2

Vértice(xv,yv)

xv=(x'+x'')/2=(0+2)/2= 1

yv=f(xv)=f(1)=-2*1²+4*1 =-2+4=2