How do you write the vertex form equation of the parabola #y=x^2-2x-5#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer salamat Apr 27, 2017 #y = (x - 1)^2 - 6# Explanation: #y = x^2 - 2 x -5# #y = (x -1)^2 -(-1)^2 -5# #y = (x - 1)^2 - 6# since a coeficient of #(x-1)^2# is positive value, it has minimum at #-6# and it axis of simetry at #x = 1#. 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 2488 views around the world You can reuse this answer Creative Commons License