Question

me ajudem por favor dadas a função F(x)=x2-7x+12 letra a suas raízes letra b valor mínimo de F(x)

Answer 1

x²-7x+12

a = 1
b = -7
c = 12

a)
Usando a resolução por Δ (delta), temos:

Δ = b²-4ac
Δ = (-7)²-4.1.12
Δ = 49-48
Δ = 1

x' = b+√Δ/2a
x' = -7+1/2
x' = -6/2
x' = -3

x'' = b-√Δ/2a
x'' = -7-1/2
x'' = -8/2
x'' = -4

Raízes: (-3; -4)

b)
Para achar o valor mínimo, temos que achar as coordenadas do Y do vértice (Yv). Para isso, usamos a fórmula:

Yv = -Δ/4a
Yv = -1/4 ou -0,25.

R: -1/4 ou -0,25.

Answer 2

$a)\\f(x)=x^2-7x+12\\\\\\0=x^2-7x+12\\\\\text{Coeficientes:}\\\\a=1\qquad\qquad{b}=-7\qquad\qquad{c}=12\\\\\mathrm{F\acute{o}rmula\ de\ Bh\acute{a}skara:}\\\\\\x=\dfrac{-b\pm\sqrt{\Delta}}{2a},\quad\mathrm{com\ }\Delta=b^2-4\cdot{a}\cdot{c}\\\\\\\text{Substituindo os valores dos coeficientes:}\\\\\\x=\dfrac{-b\pm\sqrt{b^2-4\cdot{a}\cdot{c}}}{2a}\\\\\\x=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot{1}\cdot{12}}}{2(1)}\\\\\\x=\dfrac{7\pm\sqrt{49-48}}{2}\\\\\\x=\dfrac{7\pm1}{2}\left\langle\begin{matrix}x'=4\\\\x''=3\end{matrix}\right\\\\\\\boxed{S=\{3,\ 4\}}$




$b)\\f(x)_v=-\dfrac{\Delta}{4a}\\\\\\f(x)_v=-\dfrac{(-7)^2-4\cdot{1}\cdot{12}}{4(1)}\\\\\\f(x)_v=-\dfrac{49-48}{4}\\\\\\\boxed{f(x)_v=-\dfrac{1}{4}}$