How do you find the vertex for # y=x^2-x-2#? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs 1 Answer Binayaka C. Aug 2, 2017 Vertex is at # (1/2 , -2 1/4) # Explanation: # y= x^2 -x -2 or y= x^2 -x +(1/2)^2 -1/4 -2 # or # y= (x-1/2)^2 -9/4 # , Comparing with vertex form of equation #y= a (x-h)^2+k ; (h,k)# being vertex , we find here #h=1/2 , k = -9/4# . So vertex is at # (1/2 , -2 1/4) # graph{x^2-x-2 [-10, 10, -5, 5]} [Ans] Answer link Related questions What are the important features of the graphs of quadratic functions? What do quadratic function graphs look like? How do you find the x intercepts of a quadratic function? How do you determine the vertex and direction when given a quadratic function? How do you determine the range of a quadratic function? What is the domain of quadratic functions? How do you find the maximum or minimum of quadratic functions? How do you graph #y=x^2-2x+3#? How do you know if #y=16-4x^2# opens up or down? How do you find the x-coordinate of the vertex for the graph #4x^2+16x+12=0#? See all questions in Quadratic Functions and Their Graphs Impact of this question 943 views around the world You can reuse this answer Creative Commons License