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Brunosouza203
@Brunosouza203
December 2019
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Raizes e vértice da função 2x.(5-x)=x²+3
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walacebarros
F(x) = 2x(5-x)=x²+3
f(x) = 10x - 2x²=x²+3
f(x) = 10x -2x²-x²-3=0
f(x) = -3x²+10x-3=0
coeficientes
A= -3
B= 10
C= -3
Jogando os coeficientes na formula de bhaskara, iremos encontrar as raízes da função:
x1= 3
x2= 1/3
Interseção com o eixo y = -3(0)²+10(0) -3
y = -3
vértice
xv= -b/2a
xv= 5/3
yv= -Δ/4a Δ=b²-4ac Δ=64
yv= -(64)/4(-3)
yv= 16/3
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f(x) = 10x - 2x²=x²+3
f(x) = 10x -2x²-x²-3=0
f(x) = -3x²+10x-3=0
coeficientes
A= -3
B= 10
C= -3
Jogando os coeficientes na formula de bhaskara, iremos encontrar as raízes da função:
x1= 3
x2= 1/3
Interseção com o eixo y = -3(0)²+10(0) -3
y = -3
vértice
xv= -b/2a
xv= 5/3
yv= -Δ/4a Δ=b²-4ac Δ=64
yv= -(64)/4(-3)
yv= 16/3