Question

EXERCÍCIOS PARA ENTREGAR – CARO ALUNO, APÓS
REALIZAR SUA ATIVIDADE E ENVIAR, É PRECISO MARCAR COMO
CONCLUÍDA SENÃO FICARÁ COM A SITUAÇÃO “PENDENTE”. (no
Classroom)
Determine os zeros da função quadrática
a) y = -x2 + 2x + 3
b) y = x2 – 5x + 4
c) y = x2 – 4x + 4
Encontre as coordenadas do vértice para cada função quadrática:
a) y = x2 – 6x + 8
b) y = -x2 – 10x + 11
c) y = x2 – 6x + 3
d) y = x2 – x + 1

QUEM RESPONDER CERTO ANTES DA 00:00 CHAMO NO WHATS E DOU UM PRÊMIO $$

Answer 1

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1)

$\tt{a)}~\sf{y=-x^2+2x+3}\\\sf{-x^2+2x+3=0\cdot(-1)\implies x^2-2x-3=0}\\\sf{\Delta=4+12=16}\\\sf{x=\dfrac{2\pm4}{2}}\begin{cases}\sf{x_1=3}\\\sf{x_2=-1}\end{cases}\\\tt{b)}~\sf{y=x^2-5x+4}\\\sf{x^2-5x+4=0}\\\sf{\Delta=25-16=9}\\\sf{x=\dfrac{5\pm3}{2}}\begin{cases}\sf{x_1=4}\\\sf{x_2=1}\end{cases}\\\tt{c)}~\sf{y=x^2-4x+4}\\\sf{x^2-4x+4=0\implies (x-2)^2=0}\\\sf{x-2=0}\\\sf{x=2}$

2)

$\tt{a)}~\sf{y=x^2-6x+8}\\\sf{x_V=-\dfrac{b}{2a}}\\\sf{x_V=-\dfrac{(-6)}{2\cdot1}=\dfrac{6}{2}=3}\\\sf{\Delta=36-32=4}\\\sf{y_V=-\dfrac{\Delta}{4a}}\\\sf{y_V=-\dfrac{4}{4\cdot1}}\\\sf{y_V=-\dfrac{4}{4}=-1}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{V(2,-1)}}}}}$

$\tt{b)}~\sf{y=-x^2-10x+11}\\\sf{x_V=-\dfrac{b}{2a}}\\\sf{x_V=-\dfrac{(-10)}{2\cdot(-1)}}\\\sf{x_V=\dfrac{10}{2}=5}\\\sf{\Delta=100+44=144}}\\\sf{y_V=-\dfrac{\Delta}{4a}}\\\sf{y_V=-\dfrac{144}{4\cdot(-1)}=\dfrac{144}{4}=36}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{V(5,36)}}}}}$

$\tt{c)}~\sf{y=x^2-6x+3}\\\sf{x_V=-\dfrac{b}{2a}}\\\sf{x_V=-\dfrac{(-6)}{2\cdot1}}\\\sf{x_V=\dfrac{6}{2}=3}\\\sf{\Delta=36-32=4}\\\sf{y_V=-\dfrac{\Delta}{4a}}\\\sf{y_V=-\dfrac{4}{4\cdot1}}\\\sf{y_V=-\dfrac{4}{4}}\\\sf{y_V=-1}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf{V(3,-1)}}}}}}$

$\tt{d)}~\sf{y=x^2-x+1}\\\sf{x_V=-\dfrac{-b}{2a}}\\\sf{x_V=-\dfrac{(-1)}{2\cdot1}}\\\sf{x_V=\dfrac{1}{2}}\\\sf{\Delta=1-4=-3}\\\sf{y_V=-\dfrac{\Delta}{4a}}\\\sf{y_V=-\dfrac{(-3)}{4\cdot1}}\\\sf{y_V=\dfrac{3}{4}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{V\left(\dfrac{1}{2},\dfrac{3}{4}\right)}}}}}}$

Answer 2

Explicação passo-a-passo:

1)

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

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

$\sf \Delta=2^2-4\cdot(-1)\cdot3$

$\sf \Delta=4+12$

$\sf \Delta=16$

$\sf x=\dfrac{-2\pm\sqrt{16}}{2\cdot(-1)}=\dfrac{-2\pm4}{-2}$

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

$\sf x"=\dfrac{-2-4}{-2}~\Rightarrow~x"=\dfrac{-6}{-2}~\Rightarrow~\red{x"=3}$

Os zeros dessa função são $\sf -1~e~3$

b) $\sf y=x^2-5x+4$

$\sf x^2-5x+4=0$

$\sf \Delta=(-5)^2-4\cdot1\cdot4$

$\sf \Delta=25-16$

$\sf \Delta=9$

$\sf x=\dfrac{-(-5)\pm\sqrt{9}}{2\cdot1}=\dfrac{5\pm3}{2}$

$\sf x'=\dfrac{5+3}{2}~\Rightarrow~x'=\dfrac{8}{2}~\Rightarrow~\red{x'=4}$

$\sf x"=\dfrac{5-3}{2}~\Rightarrow~x"=\dfrac{2}{2}~\Rightarrow~\red{x"=1}$

Os zeros dessa função são $\sf 1~e~4$

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

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

$\sf \Delta=(-4)^2-4\cdot1\cdot4$

$\sf \Delta=16-16$

$\sf \Delta=0$

$\sf x=\dfrac{-(-4)\pm\sqrt{0}}{2\cdot1}=\dfrac{4\pm0}{2}$

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

$\sf \red{x'=x"=2}$

O zero dessa função é $\sf 2$

2)

a) $\sf y=x^2-6x+8$

• $\sf x_V=\dfrac{-b}{2a}$

$\sf x_V=\dfrac{-(-6)}{2\cdot1}$

$\sf x_V=\dfrac{6}{2}$

$\sf \red{x_V=3}$

• $\sf y_V=\dfrac{-\Delta}{4a}$

$\sf \Delta=(-6)^2-4\cdot1\cdot8$

$\sf \Delta=36-32$

$\sf \Delta=4$

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

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

$\sf \red{y_V=-1}$

O vértice é $\sf V(3,-1)$

b) $\sf y=-x^2-10x+11$

• $\sf x_V=\dfrac{-b}{2a}$

$\sf x_V=\dfrac{-(-10)}{2\cdot(-1)}$

$\sf x_V=\dfrac{10}{-2}$

$\sf \red{x_V=-5}$

• $\sf y_V=\dfrac{-\Delta}{4a}$

$\sf \Delta=(-10)^2-4\cdot(-1)\cdot11$

$\sf \Delta=100+44$

$\sf \Delta=144$

$\sf y_V=\dfrac{-144}{4\cdot(-1)}$

$\sf y_V=\dfrac{-144}{-4}$

$\sf \red{y_V=36}$

O vértice é $\sf V(-5,36)$

c) $\sf y=x^2-6x+3$

• $\sf x_V=\dfrac{-b}{2a}$

$\sf x_V=\dfrac{-(-6)}{2\cdot1}$

$\sf x_V=\dfrac{6}{2}$

$\sf \red{x_V=3}$

• $\sf y_V=\dfrac{-\Delta}{4a}$

$\sf \Delta=(-6)^2-4\cdot1\cdot3$

$\sf \Delta=36-12$

$\sf \Delta=24$

$\sf y_V=\dfrac{-24}{4\cdot1}$

$\sf y_V=\dfrac{-24}{4}$

$\sf \red{y_V=-6}$

O vértice é $\sf V(3,-6)$

d) $\sf y=x^2-x+1$

• $\sf x_V=\dfrac{-b}{2a}$

$\sf x_V=\dfrac{-(-1)}{2\cdot1}$

$\sf \red{x_V=\dfrac{1}{2}}$

• $\sf y_V=\dfrac{-\Delta}{4a}$

$\sf \Delta=(-1)^2-4\cdot1\cdot1$

$\sf \Delta=1-4$

$\sf \Delta=-3$

$\sf y_V=\dfrac{-(-3)}{4\cdot1}$

$\sf \red{y_V=\dfrac{3}{4}}$

O vértice é $\sf V\Big(\dfrac{1}{2},\dfrac{3}{4}\Big)$