Question

f:R⇒R
f(x)=3X²-4x+1
a) intersectia cu axele OX si OY
b) coordonatele varfului
c) rezolvati ecuatia: f(x)=4x+2

Answer 1

a) intersectia cu ox  are y=0, dar y=f(x)
deci 3x²-4x+1=0
3x²-3x-x+1=0
3x(x-1)-(x-1)=0
(3x-1)(x-1)=0
x₁=1/3
x₂=1

intersectia cu 0y are x=0
y=f(0)=1

b) varful e de coordonate (-b/2a,-Δ/4a)
-b/2a=4/6=2/3
Δ=b²-4ac=16-12=4
-Δ/4a=-4/12=-1/3
deci V(2/3,-1/3)

c) 3x²-4x+1=4x+2
3x²-8x-1=0
Δ=64+12=76
x₁,₂=(4+-√76)/6

O zi buna!