Question

qual a raiz e o vértice da função y=x²-5x
 

Answer 1

$f(x)=x^2-5x$

$x^2-5x=0\\x(x-5)=0\\x'=0\\ \\x''=5$


Deriva e iguala a zero  para determinar o ponto crítico

$y=x^2-5x\\y'=2x-5\\ \\ \\0=2x-5\\ \\2x=5\\ \\x= \frac{5}{2}$

$f( \frac{5}{2} )=( \frac{5}{2} )^2- \frac{5.5}{2} \\ \\f( \frac{5}{2} )= \frac{25}{4} - \frac{25}{2} \\ \\f( \frac{5}{2} )= -\frac{25}{4}$

$\boxed{V( \frac{5}{2}~:- \frac{25}{4} )}$