Dada a função f(x) =-3(x-2)(x+1) determine os valores de
A) f(-2)
B) x tal que f(x)=0, caso existam
A) f(-2)
B) x tal que f(x)=0, caso existam
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a) f(-2) = -3 (-2-2)(-2+1)
f(-2) = -3 (-4)(-1)
f(-2) = -3 * 4
f(-2) = -12
b) f(x) = 0 = -3 (x-2)(x+1)
f(x) = 0 = -3x + 6 (x+1)
f(x) = 0 = -3x² - 3x + 6x + 6
f(x) = 0 = -3x² +3x + 6 = 0
Usando a formula de baskara...
x = -1
x = 2
f(x)=-3(x^2+x-2x-2)
f(x)= -3(x^2 -x-2)
f(x)=-3x^2+3x+6
a) f(-2)=-3(-2)^2 +3(-2)+6
f(-2)= -3*4-6+6
f(-2)= -12
b)f(x)=0
-3x^2+3x+6=0
delta=b^2 -4*a*c
delta= (3)^2 -4*(-3)*6
delta= 9+72
delta=81
-b+- raiz delta/2*a
-3+- raiz 81/2*(-3)
-3+-9/-6
x'=-3+9/-6= 6/-6=-1
x''= -3-9/-6= -12/-6=2