What is the vertex form of #y = 12x^2 -12x + 16#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer Binayaka C. Aug 25, 2016 Vertex form of equation is #y=12(x-1/2)^2+13# Explanation: #y=12x^2-12x+16=12(x^2-x)+16 =12(x^2-x+(1/2)^2)-3+16=12(x-1/2)^2+13 :.#Vertex is at #(1/2,13)#& vertex form of equation is #y=12(x-1/2)^2+13 :.# graph{12x^2-12x+16 [-80, 80, -40, 40]}[Ans] 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 1322 views around the world You can reuse this answer Creative Commons License