So we gots apolynomial equation that is 3rd degree and crosses x axis at x=-1, x=0 and x=2
the factors of a poly equation that passes through r1,r2,r3 is
(x-r1)(x-r2)(x-r3)
so
r1=-1
r2=0
r3=2
f(x)=(x-(-1))(x-0)(x-2)
f(x)=(x+1)(x)(x-2)
f(x)=x³-x²-2x
leading coefient is posiitve because 3rd degree equations that are postive go bottom left to top right
yah
actually that looks like the exact graph
both
f(x)=(x+1)(x)(x-2)
and
f(x)=x³-x²-2x
are correct
Answer:
2x^2+4x-16
Step-by-step explanation:
The quadratic can be written as
f(x) = a(x-z1)(x-z2) where z1 and z2 are the roots
f(x) = a (x-2)(x- -4)
a is the leading coefficient
f(x) = 2(x-2)(x+4)
= 2(x^2 -2x+4x-8)
= 2(x^2 +2x-8)
= 2x^2 +4x-16
Answer:
it has one solution
Step-by-step explanation:
1.y=x-3
2. 3y-3x sub x-3 in place of y therefore
it can also be written as 3x-3x-9=9
if you add 9 to both sides 3x-3x-9+9=-9+9
0+0=0
Answer:
B.) is the answer bro I got the same test bro like just trust me and give me 5 start and like and give a heart
