10-dada a funçao quadratica f(x)=x ao quadrado+5x-4,determine. a)f(1)= b)f(-2)= c)f(-4)= d)f(-6)= e)f(5)= f)f(½)= g)f(2/5)= h)f(-2/7)= (resposta em calculo) URGENTE obd
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f(x)=x^2+5x-4
a)
f(1)=1^2+5•1-4
f(1)=1+5-4=2
#f(1)=2
b)
f(-2)= (-2)^2+5•(-2)-4
f(-2)=4-10-4=-10
# f(-2)=-10
c)
f(-4)= (-4)^2+5•(-4)-4
f(-4)=16-20-4
f(-4)=-8
d)
f(-6)= (-6)^2+5•(-6)-4
f(x)=36-30-4
f(-6)=2
e)
f(5)= 5^2+5•(5)-4
f(5)=25+25-4
f(5)=46
f)
f(½)= (½)^2+5•(½)-4
f(½)= (1/4)+5/2-4# mmc= 4
f(½)=( 1+10-16)/4
f(½)= -5/4
g)
f(2/5)= (2/5)^2+5•(2/5)-4
f(2/5)= 4/25+10/5-4
f(2/5)=4/25+2-4# mmc=25
f(2/5)=(4+50-100)/25
f(2/5)=-46/25
h)
f(-2/7)= (-2/7)^2+5•(-2/7)-4
f(-2/7)= 4/49-10/7-4# mmc=49
f(-2/7)= (4-70-196)/49
f(-2/7)= -262/49