Question

Dada a função f(x)= -1x² + 1x + 2. Quanto a concavidade, determine os zeros da função e o vértice da função. (função quadrática) POR FAVOR!

Answer 1

zeros:

$\mathsf{-{x}^{2}+x+2=0}$

$\mathsf{\Delta={b}^{2}-4ac}\\\mathsf{\Delta={1}^{2}-4.(-1).2}\\\mathsf{\Delta=1+8}\\\mathsf{\Delta=9}$

$\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{x=\dfrac{-1\pm\sqrt{9}}{2.(-1)}}$

$\mathsf{x=\dfrac{-1\pm3}{-2}}\\\mathsf{x_{1}=\dfrac{-1+3}{-2}=-1}\\\mathsf{x_{2}=\dfrac{-1-3}{-2}=2}$

vértice da função

$\mathsf{x_{v}=-\dfrac{b}{2a}}\\\mathsf{x=-\dfrac{1}{2.(-1)}=\dfrac{1}{2}}$

$\mathsf{y_{v}=-\dfrac{\Delta}{4a}}\\\mathsf{y_{v}=-\dfrac{9}{4.(-1)}=\dfrac{9}{4}}$

Answer 2

f(x) = -1x² + 1x + 2

Δ = b² - 4ac

Δ = 1² - 4. (-1) .2

Δ = 1 + 8

Δ = 9

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$x' = \frac{- 1 + \sqrt{9} }{2.(-1)}$

$x' = \frac{-1+3}{-2}$

$x' = \frac{2}{-2}$

$x' = -1$

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$x'' = \frac{-1 - 3}{-2}$

$x'' = \frac{-4}{-2}$

$x'' = 2$

$As raizes:$

$S = (-1,2})$

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$Vertices:$$(\frac{1}{2} , \frac{9}{4})$

$\frac{-b}{2a}$

$\frac{-1}{2(-1)}$

$\frac{-1}{-2}$

$\frac{1}{2}$

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$\frac{-delta}{4a}$

$\frac{-9}{4(-1)}$

$\frac{-9}{-4}$

$\frac{9}{4}$