What is the vertex form of #y=2x^2 -452x-68 #? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer Archish R. Oct 14, 2017 vertex #=(113,-25606)# Explanation: #y=2x^2-452x-68# vertex form: #y=a(x-h)^2+k# where #(h,k)# is the vertex #y=2(x^2-2*113*x+12769)-25538-68# #y=2(x-113)^2-25606# #=># vertex #=(113,-25606)# 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 1349 views around the world You can reuse this answer Creative Commons License