Estude o sinal de cada função:
f(x)= 2x² - 7x + 3
f(x)= -5x² + 7x - 2
Soluções para a tarefa
Respondido por
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f(x)=2x²-7x+3
x=
x=
x=
x=
x'=
x'=3
x"=
x"=
x"=
Se x <
ou x>3 y>0 os valores de y serão positivos
Se x>
e x<3 y<0 os valores de y serão negativos
f(x)= -5x² + 7x - 2
x=
x=
x=
x'=
x'=
x'=
x"=
x"=
x"=1
Se x>
e x<1 y>0 Os valores de y serão positivos
Se x<
ou x>1 y<0 Os valores de y serão negativos
x=
x=
x=
x=
x'=
x'=3
x"=
x"=
x"=
Se x <
Se x>
f(x)= -5x² + 7x - 2
x=
x=
x=
x'=
x'=
x'=
x"=
x"=
x"=1
Se x>
Se x<
Anexos:


lucasalm:
Vou resolver a outra agora só não postei pq demora pra editar
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