What is the vertex form of #y=6x^2+17x+12 #? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer bp Feb 16, 2016 #6(x+17/32)^2 + 5277/512# This is the required vertex form. Vertex is #(-17/32, 5277/512)# Explanation: It is #y= 6(x^2 +(17x)/6) +12# = #6(x^2 + (17x)/16 + 289/1024 -289/1024)+12# = #6(x + 17/32)^2 + 12 -6(289/1024)# =#6(x+17/32)^2 + 5277/512# This is the required vertex form. Vertex is #(-17/32, 5277/512)# 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 1417 views around the world You can reuse this answer Creative Commons License