Question

Considere a função f(x) = 3x ^ 2 - 6x + 5 e determine as coordenadas do seu vértice e o seu conjunto imagem.
$3 {x}^{2} - 6x + 5$​

Answer 1

$\boxed{\begin{array}{l}\sf f(x) =ax^2+bx+c\\\sf x_V=-\dfrac{b}{2a}\\\\\sf y_V=-\dfrac{\Delta} {4a} \\\sf V\bigg(-\dfrac{b}{2a},-\dfrac{\Delta}{4a}\bigg) \longrightarrow ponto~de~m\acute inimo) (m\acute aximo) \\\sf y_V\longrightarrow valor~m\acute inimo(m\acute aximo)\\\sf se~a>0\longrightarrow admite~m\acute inimo\\\sf a<0\longrightarrow admite~m\acute aximo\\\sf para~a>0:~Imf(x) =\{y\in\mathbb{R}/y\geqslant y_V\}\\\sf se~a<0:~ Imf(x)=\{y\in\mathbb{R}/y\leqslant y_V\}\end{array}}$

$\boxed{\begin{array}{l}\sf f(x) =3x^2-6x+5\\\sf a=3>0\longrightarrow admite~m\acute inimo\\\sf\Delta= b^2-4ac\\\sf\Delta=(-6)^2-4\cdot3\cdot5\\\sf\Delta=36-60\\\sf\Delta=-24\\\sf x_V=-\dfrac{b}{2a}\\\\\sf x_V=-\dfrac{-6}{2\cdot3}=1\\\\\sf y_V=-\dfrac{\Delta}{4a}\\\\\sf y_V=-\dfrac{-24}{4\cdot3}=2\\\sf V(1,2)\\\sf Im=\{x\in\mathbb{R}/x\geqslant2\} \end{array}}$