How do you write the vertex form equation of the parabola #y=2x^2 +11x-6#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer EZ as pi May 10, 2016 # y =2(x + 11/4)^2 - 21 1/8# Explanation: # y = 2x^2 + 11x - 6" divide RHS by 2"# # y = 2(x^2 + 11/2 " " - 3 " ") " leave spaces to add terms"# # y = 2(x^2 + 11/2 + color(red)((11/4)^2) - 3 - color(red)( (11/4)^2))# #" " 11/2 ÷ 2 = 11/4# #y = 2[(x + 11/4)^2 -169/16]# #y =2(x + 11/4)^2 - 2xx169/16# #y =2(x + 11/4)^2 - 169/8 rArr y =2(x + 11/4)^2 - 21 1/8# Answer link Related questions What is the Vertex Form of a Quadratic Equation? How do you find the vertex form of a quadratic equation? How do you graph quadratic equations written in vertex form? How do you write #y+1=-2x^2-x# in the vertex form? How do you write the quadratic equation given #a=-2# and the vertex #(-5, 0)#? What is the quadratic equation containing (5, 2) and vertex (1, –2)? How do you find the vertex, x-intercept, y-intercept, and graph the equation #y=-4x^2+20x-24#? How do you write #y=9x^2+3x-10# in vertex form? What is the vertex of #y=-1/2(x-4)^2-7#? What is the vertex form of #y=x^2-6x+6#? See all questions in Vertex Form of a Quadratic Equation Impact of this question 964 views around the world You can reuse this answer Creative Commons License