I don't know to find parabola mx^2+2mx+x+m-1. all i know is the extreme of parabola is on the line 9x-3y-3.

To determine the parabola, we'll have to find out the
unknown m.

The extreme point of the parabola is the vertex
of the parabola.

If the vertex of the parabola belongs to
the line 9x-3y-3 = 0, then the coordinates of the vertex verify the equation of the
line.

Let's note the vertex as V(xV,
yV)

xV = -b/2a

yV =
-delta/4a

We'll substitute xV and yV in the equation of the
line:

9xV - 3yV - 3 = 0

We'll
divide by 3:

3xV - yV - 1 =
0

We'll put the equation in the slope intercept
form:

yV = 3xV - 1

-delta/4a =
-3b/2a - 1

delta = b^2 -
4ac

We'll identify a,b,c:

a =
m

b = 2m+1

c =
m-1

delta = (2m+1)^2 -
4m(m-1)

We'll expand the square and we'll remove the
brackets:

4m^2 + 4m + 1 - 4m^2 +
4m

We'll eliminate like
terms:

delta = 8m +
1

-delta/4a = -3b/2a -
1

-(8m+1)/4m = -3(2m+1)/2m -
1

-8m-1 = -12m - 6 - 4m

8m + 5
= 0

We'll add -5 both
sides:

8m = -5 

We'll divide
by 8:

m =
-5/8

The parabola is: -5x^2/8 - x/4 -
13/8