Question

Esboce o gráfico: f(x)= x²-8x+12

Answer 1

Segue esboço do gráfico em anexo.

$f(x)=x^2-8x+12$


$\bullet\;\;$ Interseção com o eixo $y:$

Fazendo $x=0\,,$

$f(x)=12$

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$\bullet\;\;$ Interseção com o eixo $x:$

Encontrando as raízes da equação:

$f(x)=0\\\\ x^2-8x+12=0~~~\Rightarrow~~\left\{\! \begin{array}{l}a=1\\b=-8\\c=12 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=(-8)^2-4\cdot 1\cdot 12\\\\ \Delta=64-48\\\\ \Delta=16$


$x=\dfrac{-(-8)\pm \sqrt{16}}{2\cdot 1}\\\\\\ x=\dfrac{8\pm 4}{2}\\\\\\ \begin{array}{rcl} x_1=\dfrac{8-4}{2}&~\text{ e }~&x_2=\dfrac{8+4}{2}\\\\\\ x_1=\dfrac{4}{2}&~\text{ e }~&x_2=\dfrac{12}{2}\\\\\\ x_1=2&~\text{ e }~&x_2=6 \end{array}$

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$\bullet\;\;$ Coordenadas do vértice:

$\bullet\;\;x_{_V}=-\,\dfrac{b}{2a}\\\\\\ x_{_V}=-\,\dfrac{(-8)}{2\cdot 1}\\\\\\ x_{_V}=\dfrac{8}{2}\\\\\\ \boxed{\begin{array}{c} x_{_V}=4 \end{array}}$

$\bullet\;\;y_{_V}=-\,\dfrac{\Delta}{4a}\\\\\\ y_{_V}=-\,\dfrac{16}{4\cdot 1}\\\\\\ y_{_V}=-\,\dfrac{16}{4}\\\\\\ \boxed{\begin{array}{c} y_{_V}=-4 \end{array}}$


Bons estudos! :-)