A reta f(x)= -x corta a parábola f(x)= -2x^2 -2x + 3nos pontos: (A) (0,0) e (1,-1 ) (B) (1,-1) e (-3.... Pergunta de ideia deDaninovelini

Daninovelini @Daninovelini

January 2020 1 46 Report

A reta f(x)= -x corta a parábola f(x)= -2x^2 -2x + 3nos pontos: (A) (0,0) e (1,-1 ) (B) (1,-1) e (-3/2) (c) (-3/2) e (0,0) (D) (0,-1) e (0, -3/2)

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MATHSPHIS Basta igualar:

$-x=-2x^2 -2x + 3\\
\\
2x^2+x-3=0\\
\\
\Delta=1^2-4.2.(-3)\\
\\
\Delta=1+24\\
\\
\Delta=25\\
\\
x=\frac{-1\pm\sqrt{25}}{4}=\frac{-1\pm5}{4}\\
\\
x_1=-\frac{6}{4}=-\frac{3}{2}\\
\\
x_2=\frac{4}{4}=1$

$Se \ x=-\frac{3}{2} \Rightarrow y = \frac{3}{2} \ \ P_1=\left(-\frac{3}{2};\frac{3}{2}\right)
\\
\\
Se \ x=1 \Rightarrow y=-1 \ \ P_2=(1;-1)$

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