Question

Alguém consegue resolver a número 1. Preciso urgente!

Answer 1

1) 

a) $-2x^2+7x-3=0$
$a=-2~~~~~~~~~~~~~~~~~b=7~~~~~~~~~~~~~~~~c=-3$

$\Delta=b^2-4\cdot a\cdot c$
$\Delta=7^2-4\cdot(-2)\cdot(-3)$
$\Delta=49-24$
$\Delta=25$

$x=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$x=\dfrac{-7\pm\sqrt{25}}{2\cdot(-2)}=\dfrac{-7\pm5}{-4}$

$x'=\dfrac{-7+5}{-4}~\longrightarrow~x'=\dfrac{-2}{-4}~\longrightarrow~\boxed{x'=\dfrac{1}{2}}~\longrightarrow~\text{Ponto}\left(\dfrac{1}{2},0\right)$

$x"=\dfrac{-7-5}{-4}~\longrightarriw~x"=\dfrac{-12}{-4}~\longrightarrow~\boxed{x"=3}~\longrightarrow~\text{Ponto}(3,0)$


b) $\bullet~~x_{\text{V}}=\dfrac{-b}{2a}$

$x_{\text{V}}=\dfrac{-7}{2\cdot(-2)}~\longrightarrow~x_{\text{V}}=\dfrac{-7}{-4}~\longrightarrow~\boxed{x_{\text{V}}=\dfrac{7}{4}}$

$\bullet~~y_{\text{V}}=\dfrac{-\Delta}{4a}$

$y_{\text{V}}=\dfrac{-25}{4\cdot(-2)}~\longrightarrow~y_{\text{V}}=\dfrac{-25}{-8}~\longrightarrow~\boxed{y_{\text{V}}=\dfrac{25}{8}}$

$\text{V}\left(\dfrac{7}{4},\dfrac{25}{8}\right)$

c) $a<0~\longrightarrow~\text{Valor}~\text{M}\acute{a}\text{ximo}$

d) $x=0~\longrightarrow~y=-2\cdot0^2+7\cdot0-3~\longrightarrow~y=-3~~~\boxed{\text{Ponto}(0,-3)}$ 

e) Como $\text{y}_{\text{V}}=\dfrac{25}{8}$ é o valor máximo, temos que $\text{Im}(y)=\left\{y\in\mathbb{R}~|~y\le\dfrac{25}{8}\right\}$

Veja o gráfico na imagem em anexo.