# If the vertex of the curve ax^2 + (2a+1)x + a-1 is on the line y=3x-1, which is a?

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Let f(x) = ax^2 + (2a+1)x + a -1

First let us dtermine the vertex:

V(xv, yv) such that:

==> xv = -b/2a = -2a-1/2a = -1 - 1/2a

==> yv = -(b^2 -4ac)/4a = [-(2a+1)^2 + 4a(a-1)]/4a

= [-4a^2 -4a-1 + 4a^2 - 4a]/4a

= -8a-1/4a

==> yv = -2a -1/4a

But the vertext V is on the line : y= 3x-1

Then the point V should verify the point.

==> y= 3x -1

==>( -8a -1)/4a = 3[(-2a - 1)/2a] -1

==> (-8a-1) = 6(-2a-1) - 4a

==> -8a -1 = -12a -6 -4a

==> 8a = -5

**==> a= -5/8**

If the vertex of the parable belongs to the line y=3x-1, 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:

yV = 3xV - 1

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

delta = b^2 - 4ac

We'll identify a,b,c:

a = a

b = 2a+1

c = a-1

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

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

4a^2 + 4a + 1 - 4a^2 + 4a

We'll eliminate like terms:

delta = 8a + 1

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

-(8a+1)/4a = -3b/2a - 1

-8a-1 = -6a - 4a

-8a + 10a - 1 = 0

We'll combine like terms:

2a - 1 = 0

We'll add 1 both sides:

2a = 1

We'll divide by 2:

**a = 1/2**