Question

Determine m, sendo m pertence a R, de modo que o valor mínimo da função f(x) = x² + 4x + m seja 1, é: * 1 ponto m = 2 m = 5 m = -5 m = -2 m = 4​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf func_{\!\!,}\tilde ao\,quadr\acute atica}\\\rm f(x)=ax^2+bx+c\,onde\,a,b,c\in\mathbb{R}\,e\,a\ne0\\\rm se\,a>0\longrightarrow admite\,m\acute inimo\\\rm se~a<0\longrightarrow admite\,m\acute aximo\\\rm o\,valor\,m\acute inimo(m\acute aximo)\,da\,func_{\!\!,}\tilde ao\\\rm \acute e\,obtido\,pela\,express\tilde ao\,y_V=-\dfrac{\Delta}{4a}\\\rm onde\,\Delta=b^2-4ac\end{array}}$

$\large\boxed{\begin{array}{l}\rm f(x)=x^2+4x+m\\\rm a=1>0\longrightarrow admite\,m\acute inimo\\\rm\Delta=b^2-4ac\\\rm\Delta=4^2-4\cdot1\cdot m\\\rm\Delta=16-4m\\\rm y_V=-\dfrac{\Delta}{4a}\\\\\rm 1=-\dfrac{16-4m}{4\cdot1}\\\\\rm \dfrac{4m-16}{4}=1\\\\\rm 4m-16=4\\\rm 4m=16+4\\\rm 4m=20\\\rm m=\dfrac{20}{4}\\\\\rm m=5\end{array}}$