Question

1. Determine men para que o vértice da parábola de equação y = x2 - mx + n seja (-1,2). Classifique o vértice em ponto de máximo ou ponto de mínimo. ​

Answer 1

$\large\boxed{\begin{array}{l}\sf y=x^2-mx+n\\\sf a=1>0\longrightarrow Apresenta\,ponto\,de\,m\acute inimo\,em\,V(x_V,y_V).\\\sf \Delta=b^2-4ac\\\sf\Delta=(-m)^2-4\cdot1\cdot n\\\sf\Delta=m^2-4n\\\sf x_V=-\dfrac{b}{2a}\\\\\sf x_V=-\dfrac{-m}{2\cdot1}=\dfrac{m}{2}\\\\\sf y_V=-\dfrac{\Delta}{4a}\\\\\sf y_V=-\dfrac{m^2-4n}{4}=\dfrac{4n-m^2}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\sf V(x_V,y_V)\\\sf V\bigg(\dfrac{m}{2},\dfrac{4n-m^2}{4}\bigg)\\\sf mas~V(-1,2).\\\\\sf \dfrac{m}{2}=-1\\\sf m=2\cdot(-1)=-2\\\sf \dfrac{4n-m^2}{4}=2\\\\\sf 4n-m^2=8\\\sf 4n-(-2)^2=8\\\sf 4n-4=8\\\sf 4n=8+4\\\sf 4n=12\\\sf n=\dfrac{12}{4}\\\\\sf n=3\end{array}}$