Question

Se f (x) = 2 - x^2 , então o valor máximo
assumido por g(x) = f (x -4) + 1 é ?

Answer 1

Boa noite!


Solução!


$f(x)=2- x^{2} \\\\\\\ g(x)=f(x-4)+1\\\\\\\ g(x)=2-(x-4)^{2}+1\\\\\\\ g(x)=2-( x^{2} -4x-4x+16)+1\\\\\\ g(x)=2-( x^{2} -8x+16)+1\\\\ g(x)=2- x^{2} +8x-16+1\\\\\\ g(x)=- x^{2} +8x-16+1+2\\\\\\ g(x)=- x^{2} +8x-13$


$Altura~~maxima~~de~~g(x)~~e~~dada~~pela~~formula.$




$y_{V}= \dfrac{b^{2}-4.a.c }{-4.a}\\\\\\\ y_{V}= \dfrac{(8)^{2}-4.(-1).(-13) }{-4.(-1)}\\\\\\\ y_{V}= \dfrac{64-52 }{4}\\\\\\\ y_{V}= \dfrac{12 }{4}\\\\\\\ y_{V}=3\\\\\\\\\\ \boxed{Resposta:Valor~~maximo~~3}$

Boa noite!
Bons estudos!

Answer 2

Oi Tjk

f(x) = 2 - x² 

g(x) = f(x - 4) + 1

g(x) = 2 - (x - 4)² + 1 = 2 - x² + 8x - 16 + 1 = -x² + 8x - 13

delta
d² = 8² - 4*(-1)*(-13) = 64 - 52 = 12

vértice 

Vy = -d²/4a = -12/-4 = 3 o valor máximo pedido