What is the vertex form of #y= (x -4)(x/2 - 2) #? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer Binayaka C. Jun 4, 2016 In vertex form #y=1/2(x-4)^2# Explanation: #y=(x-4)(x/2-2) = (x-4)*1/2(x-4) =1/2(x-4)^2# Vertex is at #4,0# graph{1/2(x-4)^2 [-10, 10, -5, 5]}[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 1303 views around the world You can reuse this answer Creative Commons License