Question

Gente por favor é pra hoje esta valendo nota preciso muito de ajuda !!!!!

De as coordenadas dos vértices e as raízes se existirem

a_Y=x²-4x+3
b_Y=x²-7x+10
c_Y=x²-5x+6
d_Y=x²+3x+3
e_Y=x²+4x+3
f_Y=x²-2x-3
g_Y=x²+2x-8
h_Y=x²-8x-9
i_Y=x²+11x+28
j_Y=x²+10x-24

Answer 1

        Thais,
  Muitas perguntas numa tarefa só

  Todas funções quadráticas completas
  O procedimento de solução é o mesmo
  Vou resolver duas para você conhecer a metodologia de solução.
  Com essa base, as outras levam poucos minutos
 
  As raízes de uma função quadrática são os valores da incógnita quando a
  função e nula

          a)
                  y = x² - 4x + 3    
                          x² - 4x + 3 = 0
              RAIZES
                  Fatorando
                          (x - 3)(x - 1) = 0
                                   x - 3 = 0
                                                      x1 = 3
                                   x - 1 = 0
                                                     x2 = 1
                                                                    S = { 1, 3 }
            COORDENADAS VÉRTICE
                             xV = - b/2a
                                  = - (-4)/2(1)
                                  = 4/2                   xV = 2
                             yV = - Δ/4a
                                       Δ = b² - 4.a.c
                                           = (-4)² - 4(1)(3)
                                           = 16 - 12
                                           = 4
                             yV = - 4/4(1)            yV = - 1
                                                                               Pv(2, - 1)


                     g)
                            y = x² + 2x - 8
                                  x² + 2x - 8 = 0
                  RAIZES
                      fórmula geral
                             x = (- b +/- √Δ)/2a
                                    Δ = (2)² - 4(1)(-8)
                                        = 4 + 32
                                        = 36
                                                      √36 = 6
                            x = (- 2 +/- 6)/2
                                   x = (- 2 - 6)/2
                                                             x1 = - 4
                                  x = (- 2 + 6)/2
                                                             x2 = 2
                                                                           S = { - 4, 2 }
              COORDENADAS VÉRTICE
                             xV = - 2/2
                                                     xV = - 1
                             yV = - 36/4
                                                     yV = - 9
                                                                       Pv(- 1, - 9)

Answer 2

Aqui vão mais duas:

Fórmulas ==> Por Baskara, calcular os zeros:

$\Delta=b^2-4ac~~~~e~~~~x= \dfrac{-b\pm \sqrt{\Delta} }{2a}$

Calcular o vértice:

$x_v= \dfrac{-b}{2a} ~~~~e~~~~y_v= \dfrac{-\Delta}{4a}$



$B)~~y=x^2-7x+10\to\\\\ x^2-7x+10=0\\\\ a=1;~b=-7;~c=10\\\\\\ \Delta=(-7)^2-4.1.10\to~~ \Delta=49-40\to \boxed{\Delta=9}\\\\ x' \neq x''\\\\ Dois~~zeros~~reais~~distintos: \\\\ x= \dfrac{-(-7)\pm \sqrt{9} }{2.1}\to~~x= \dfrac{7\pm 3 }{2}\to\\\\\\ x'= \dfrac{7+3 }{2}\to ~~x'= \dfrac{10}{2}\to~~ \boxed{x'=5}\\\\\\x''= \dfrac{7-3 }{2}\to ~~x''= \dfrac{4}{2}\to~~ \boxed{x''=2}\\ \large\boxed{\boxed{S=\{2~;~5\}}}$


Vértice:
$x_v= \dfrac{-(-7)}{2.1}\to~~\boxed{x_v= \dfrac{7}{2}}\\\\\\y_v= \dfrac{-9}{4.1}\to~~\boxed{y_v= \dfrac{-9}{4}}\\\\\\ \large\boxed{\boxed{V=\left\{ \dfrac{7}{2}~; ~\dfrac{-9}{4} \right\}}}}$






$F)~~y=x^2-2x-3\to \\x^2-2x-3=0\\ a=1;~b=-2;~c=-3\\\Delta=(-2)^2-4.1.(-3)\to~~\Delta=4+12\to~~\boxed{\Delta=16}\\ x' \neq x''\\ Dois~~zeros ~~reais~~distintos:\\\\ x= \dfrac{-(-2)\pm \sqrt{16} }{2.1}\to~~x= \dfrac{2\pm 4 }{2}\to\\\\\\ x'= \dfrac{2+ 4 }{2}\to~~ x'= \dfrac{6 }{2}\to~~\boxed{x'=3}\\\\\\x''= \dfrac{2- 4 }{2}\to~~ x''= \dfrac{-2 }{2}\to~~\boxed{x'=-1}\\\\\\ \large\boxed{\boxed{S=\{-1~;~3\}}}$


Vértice:
$x_v= \dfrac{-(-2)}{2.1}\to~~x_v= \dfrac{2}{2}\to~~\boxed{x_v= 1}\\\\\\y_v= \dfrac{-16}{4.1}\to~~y_v= \dfrac{-16}{4}\to~~ \boxed{y_v= -4}\\\\\\ \large\boxed{\boxed{V=\left\{ 1~; ~-4 \right\}}}}$