Hi

 Parabola with x-intercept of (2,0) and a vertex at (6,8).  

Using the vertex form of a parabola,  where(h,k) is the vertex

writing in the vertex form:

 y = a(x-6)^2 + 8  Using x-intercept Pt(2,0)to solve for a

 0 = a(-4)^2 +8

-8= 16a

-1/2 = a

y = (-1/2)(x-6)^2 + 8   written in the vertex form

Standard form a parabola is y = ax^2 + bx + c

y = (-1/2)[(x-6)^2 - 16]

y = -(1/2[x^2 - 12x + 36 - 16]

y = -(1/2)x^2-12x + 20 



solving

[(3)/(3-x)] + [(4)/x+2)]=4



-3(x+2) + 4(x-3) = 4(x+2)(x-3)= 4(x^2 - x - 6)

-3x - 6 + 4x - 12 = 4x^2 - 4x - 24

  4x^2 -5x - 6 = 0

   x = 2 and x = -6/8 = -3/4

 

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