y=3x？ +2x- 1

Y=0 on the x axis.

3x？ +2x- 1=0

(3x- 1)(x+ 1)=0

x= 1/3，x=- 1

So (1/3,0), (-1, 0)

a=b= 1

y=3x？ +2x+c

- 1 & lt； X< when 1, equation 3x? +2x+c=0 has only one solution.

If the equation itself has only one solution, the discriminant is 0.

Then 4- 12c=0 and c= 1/3.

3x？ +2x+ 1/3=0

9x？ +6x+ 1=0

(3x+ 1)？ =0

X=- 1/3, which accords with-1

If the equation itself has two solutions, the discriminant is greater than 0.

Then 4-12c >; 0，c & lt 1/3

At this point, there is only one solution in-1

So-1

Then when x=- 1 and x= 1, the function value is positive and negative, that is, the multiplication is less than 0.

x= 1，y=5+c

x=- 1，y= 1+c

So (5+c) (1+c) < 0.

-5 & lt； c & lt- 1

To sum it up

-5 & lt； c & lt- 1，c= 1/3

x 1=0，y 1 = 0+0+c & gt； 0

x2= 1，y2 = 3a+2 b+ c & gt； 0

a+b+c=0

So c =-a-b >; 0

a+b & lt； 0

B<［ [ancient or Latin modern names of animals and plants]

C=-a-b substitute 3a+2 b+ c >；; 0

2a+b & gt； 0

b & gt-2a

So-2a

Namely -2a

So-3a

So-3a

So by b & gt-3a, both sides are divided by -3a.

So b/(-3a) < 1

-2a

Because of a>0, so -a

So b/(-3a) > 0

So 0

Because y=3ax? The symmetry axis of +2bx+c is x=-b/3a.

So the symmetry axis is 0.

Discriminant =4b? - 12ac

b=-a-c

So discriminant =4a? +8ac+4c？ - 12ac=4a？ -4ac+4c？ =4(a？ -ac+c？ )

=4[(a-c/2)？ +3c？ /4)]

So the discriminant is greater than or equal to 0, so the y axis and the x axis intersect.

Because of a>0, the opening is upward, so when x=-b/3a, the minimum value of y is less than or equal to 0.

And x 1 = 0, y1>; 0

x2= 1，y2 & gt0

The symmetry axis is 0.

So when 0