Question

Dada a função f(x) = x² – 8x + 8, determine: a) Coordenadas do vértice. b) Zeros da função

Answer 1

Resposta:

Vamos lá!

Zeros da função:

x' = 4+2√2

x'' = 4-2√2

x² – 8x + 8

Δ = (- 8)² - 4 . 1 . 8

Δ = 64 - 32

Δ = 32

x = -[- 8 ± √32] / 2

x = [8 ± 4√2] / 2

x' = 4+2√2

x'' = 4-2√2

Coordenadas do vértice:

$x_{v} =-\dfrac{b}{2a} \\\\x_{v} =-\dfrac{-8}{2} \\\\x_{v} =4\\\\\\y_{v} =-\dfrac{32}{4a}\\\\y_{v} =8$

f(x) = x² – 8x + 8

f(-1) = (-1)² – 8.(-1) + 8

f(-1) = 1 + 8 + 8

f(-1) = 17

f(x) = x² – 8x + 8

f(0) = 0² – 8.0 + 8

f(0) = 0 – 0 + 8

f(0) = 8  Coordenada do Vértice

f(x) = x² – 8x + 8

f(1) = 1² – 8.1 + 8

f(1) = 1 – 8 + 8

f(1) = 1

f(x) = x² – 8x + 8

f(2) = 2² – 8.2 + 8

f(2) = 4 – 16 + 8

f(2) = - 4

f(x) = x² – 8x + 8

f(3) = 3² – 8.3 + 8

f(3) = 9 – 24 + 8

f(3) = - 7

f(x) = x² – 8x + 8

f(4) = 4² – 8.4 + 8

f(4) = 16 – 32 + 8

f(4) = - 8 Coordenada do Vértice

f(x) = x² – 8x + 8

f(5) = 5² – 8.5 + 8

f(5) = 25 – 40 + 8

f(5) = - 7