Question

Analise a função quadrática f(x) = x² - 26x + 168, podemos afirmar que seu valor mínimo é​

Answer 1

Podemos fazer de duas formas :

1ª y do vértice

$\displaystyle \text{f(x)}=\text x^2-26\text x+168 \\\\ \text Y_\text v = \frac{-\Delta }{4\text a} \\\\\\ \text Y_\text v= \frac{-[(-26)^2-4.168])}{4.1} \\\\\ \text Y_\text v = \frac{-(676-672)}{4} \\\\\\ \text Y_\text v = \frac{-4}{4} \\\\\\ \text Y_\text v =-1 \\\\ \huge\boxed{\text{Valor m{\'i}nimo} = -1 \ }\checkmark$

2ª derivando igual a 0 :

$\text{f(x)}=\text x^2-26\text x+168 \\\\ \underline{\text{Derivando}}: \\\\ 2\text x-26 = 0 \\\\ \text x = 13 \\\\ \text{f(13)} = 13^2-26.13+168 \\\\ \text{f(13)}=169+168-338 \\\\ \text{f(13)}=-1 \\\\ \huge\boxed{\text{Valor m{\'i}nimo}=-1\ }\checkmark$