Question

Traçar o gráfico da função y=x²-4x​

Answer 1

Explicação passo-a-passo:

tá aí o quadro.......só não sei se está certo

Answer 2

$\large\boxed{\begin{array}{l}\underline{\rm Func_{\!\!,}\tilde ao\,do\,2^o\,grau}\\\sf \acute E\,toda\,func_{\!\!,}\tilde ao\,que\,assume\\\sf a\,forma\,f(x)=ax^2+bx+c\,onde\,a,b,c\in\mathbb{R},a\ne0\\\sf o\,gr\acute afico\,\acute e\,uma\,curva\,denominada\,par\acute abola.\\\sf se\,a>0:\\\sf concavidade\,para\,cima\,admite\,valor\,m\acute inimo\\\sf se\,a<0:\\\sf concavidade\,para\,baixo\,admite\,m\acute aximo\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Construc_{\!\!,}\tilde ao\,do\,gr\acute afico.}\\\sf 1\, determinar\,a\,intersecc_{\!\!,}\tilde ao\,da\,par\acute a bola\,com\,eixo\,y:\\\underline{\rm fac_{\!\!,}a\,x=0\,e\,calcule\,y}\\\sf 2\,determinar\,a\,intersecc_{\!\!,}\tilde ao\,da\,par\acute abola\,com\,eixo\,x:\\\underline{\rm fac_{\!\!,}a\,y=0\,e\,calcule\,x}\\\sf 3\,determinar\,as\,coordenadas\,do\,v\acute ertice\,v(x_V,y_V)\\\sf onde\, x_V=-\dfrac{b}{2a}\,e~y_V=-\dfrac{\Delta}{4a}.\end{array}}$

$\boxed{\begin{array}{l}\sf y=x^2-4x\\\sf a=1>0\implies concavidade\,para\,cima\,(admite\,m\acute inimo)\\\sf \Delta=b^2-4ac\\\sf\Delta=(-4)^2-4\cdot1\cdot0\\\sf\Delta=16\\\sf quando\,x=0:\\\sf f(0)=0^2-4\cdot0=0\implies A(0,0)\\\sf quando\,y=0:\\\sf x^2-4x=0\\\sf x\cdot(x-4)=0\\\sf x=0\\\sf x-4=0\\\sf x=4\\\sf B(0,0)\,C(4,0)\\\sf coordenadas\,do\,v\acute ertice:\\\sf x_V=-\dfrac{b}{2a}\\\\\sf x_V=-\dfrac{-4}{2\cdot1}=2\\\\\sf y_V=-\dfrac{\Delta}{4a}\\\\\sf y_V=-\dfrac{16}{4\cdot1}=-4~~V(2,-4)\end{array}}$

$\large\boxed{\begin{array}{l}\rm O\,gr\acute afico\,est\acute a\,anexo\end{array}}$