SOLUTION: 1) Write the equation of the parabola in Standard Form that an x-intercept of (2,0) and a vertex at (6,8). Not sure how to set up:( 2)Solve for x: [(3)/(3-x)] + [(4)/x+2)]=4.

Algebra ->  Algebra  -> Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: 1) Write the equation of the parabola in Standard Form that an x-intercept of (2,0) and a vertex at (6,8). Not sure how to set up:( 2)Solve for x: [(3)/(3-x)] + [(4)/x+2)]=4.       Log On

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 Click here to see ALL problems on Quadratic Equations Question 391829: 1) Write the equation of the parabola in Standard Form that an x-intercept of (2,0) and a vertex at (6,8). Not sure how to set up:( 2)Solve for x: [(3)/(3-x)] + [(4)/x+2)]=4. I think -3+-sqrt681/-8 ?? Answer by ewatrrr(10682)   (Show Source): You can put this solution on YOUR website!``` 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 ```