Resposta :
Olá, Vinícius.
[tex]a) f(x)=x\²-2x-3\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{-2}2=1\\y_v=f(x_v)=1-2-3=-4\end{cases} \\\\\\ b) f(x)=-x\²+3x-5\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{3}{-2}=\frac32\\y_v=f(x_v)=-\frac94+\frac92-5=\frac{-9+18-20}4=-\frac{11}4\end{cases} \\\\\\ c) f(x)=x\²-4x+3\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{-4}2=2\\y_v=f(x_v)=4-8+3=-1\end{cases}[/tex]
[tex]a) f(x)=x\²-2x-3\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{-2}2=1\\y_v=f(x_v)=1-2-3=-4\end{cases} \\\\\\ b) f(x)=-x\²+3x-5\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{3}{-2}=\frac32\\y_v=f(x_v)=-\frac94+\frac92-5=\frac{-9+18-20}4=-\frac{11}4\end{cases} \\\\\\ c) f(x)=x\²-4x+3\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{-4}2=2\\y_v=f(x_v)=4-8+3=-1\end{cases}[/tex]
Resposta:
[tex]a) f(x)=x\²-2x-3\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{-2}2=1\\y_v=f(x_v)=1-2-3=-4\end{cases} \\\\\\ b) f(x)=-x\²+3x-5\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{3}{-2}=\frac32\\y_v=f(x_v)=-\frac94+\frac92-5=\frac{-9+18-20}4=-\frac{11}4\end{cases} \\\\\\ c) f(x)=x\²-4x+3\Rightarrow\begin{cases}x_v=-\frac b{2a}=-\frac{-4}2=2\\y_v=f(x_v)=4-8+3=-1\end{cases}[/tex]
Explicação passo-a-passo: