Question

Construir o grafico das funções quadraticas

y=2x²-4x+1

Answer 1

$\large\begin{array}{l} \textsf{Construir o gr\'afico da fun\c{c}\~ao do segundo grau, cuja lei \'e}\\\\ \mathsf{y=2x^2-4x+1}\\\\\\\textsf{(o gr\'afico segue em anexo)} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Interse\c{c}\~ao com o eixo y.}\\\\ \textsf{Fazendo }\mathsf{x=0,}\textsf{ encontramos}\\\\ \mathsf{y=2\cdot 0^2-4\cdot 0+1}\\\\ \mathsf{y=1}\\\\\\ \textsf{O gr\'afico da fun\c{c}\~ao interseciona o eixo y no ponto (0,\,1).} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Interse\c{c}\~oes com o eixo x (ra\'izes da fun\c{c}\~ao).}\\\\ \textsf{Fazendo }\mathsf{y=0,}\\\\ \mathsf{2x^2-4x+1=0}\quad\Rightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\\mathsf{b=-4}\\\mathsf{c=1} \end{array} \right. \end{array}$


$\large\begin{array}{l} \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(-4)^2-4\cdot 2\cdot 1}\\\\ \mathsf{\Delta=16-8}\\\\ \mathsf{\Delta=8}\\\\ \mathsf{\Delta=2^2\cdot 2} \end{array}$


$\large\begin{array}{l} \mathsf{x=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{x=\dfrac{-(-4)\pm \sqrt{2^2\cdot 2}}{2\cdot 2}}\\\\ \mathsf{x=\dfrac{4\pm 2\sqrt{2}}{2\cdot 2}}\\\\ \mathsf{x=\dfrac{\diagup\!\!\!\! 2\cdot (2\pm \sqrt{2})}{\diagup\!\!\!\! 2\cdot 2}}\\\\ \mathsf{x=\dfrac{2\pm \sqrt{2}}{2}}\\\\ \mathsf{x=1\pm \dfrac{\sqrt{2}}{2}} \end{array}$

$\large\begin{array}{l} \mathsf{x_1=1-\dfrac{\sqrt{2}}{2}~~~e~~~x_2=1+\dfrac{\sqrt{2}}{2}}\\\\\\ \textsf{O gr\'afico interseciona o eixo x nos pontos }\\\\\mathsf{(1-\frac{\sqrt{2}}{2},\,0)~~e~~(1+\frac{\sqrt{2}}{2},\,0).} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{As coordenadas do v\'ertice }\mathsf{(x_{_V},\,y_{_V}).}\\\\ \mathsf{x_{_V}=-\,\dfrac{b}{2a}}\\\\ \mathsf{x_{_V}=-\,\dfrac{(-4)}{2\cdot 2}}\\\\ \mathsf{x_{_V}=\dfrac{4}{4}}\\\\ \mathsf{x_{_V}=1} \end{array}$


$\large\begin{array}{l} \mathsf{y_{_V}=-\,\dfrac{8}{4\cdot 2}}\\\\ \mathsf{y_{_V}=-\,\dfrac{8}{8}}\\\\ \mathsf{y_{_V}=-1}\\\\\\\textsf{O v\'ertice \'e o ponto }\mathsf{(1,\,-1).} \end{array}$


$\large\begin{array}{l} \textsf{Como o coeficiente }\mathsf{a=2}\textsf{ \'e positivo, o gr\'afico ser\'a uma}\\\textsf{par\'abola com concavidade voltada para cima, e a fun\c{c}\~ao}\\\textsf{possui um ponto de m\'inimo em }\mathsf{x=x_{_V}=1.} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Calculando o valor da fun\c{c}\~ao quando }\mathsf{x=2x_{_V}.}\\\\ \textsf{Este passo \'e importante para que o esbo\c{c}o do gr\'afico fique}\\\textsf{melhor, visto que o gr\'afico possui um eixo de simetria em}\\\mathsf{x=x_{_V}.}\\\\ \textsf{Quando }\mathsf{x=2x_{_V},}\textsf{ temos}\\\\ \mathsf{x=2\cdot 1}\\\\ \mathsf{x=2}\\\\\\ \mathsf{y=2\cdot 2^2-4\cdot 2+1}\\\\ \mathsf{y=2\cdot 4-8+1}\\\\ \mathsf{y=8-8+1}\\\\ \mathsf{y=1}\\\\\\ \textsf{O ponto (2,\,1) pertence ao gr\'afico da fun\c{c}\~ao.} \end{array}$


$\large\begin{array}{l} \textsf{Agora, \'e s\'o marcar os pontos encontrados no plano cartesiano}\\\textsf{e esbo\c{c}ar o gr\'afico. Ser\'a uma par\'abola passando por estes}\\\textsf{pontos.} \end{array}$


Caso tenha problemas para visualizar a resposta, experimente abrir pelo navegador: http://brainly.com.br/tarefa/7381033


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: esboço gráfico função quadrática segundo grau