What is the vertex of # y= 2x^2-x-3-4(x-1)^2#? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs 1 Answer mizoo May 1, 2018 #"Vertex": (7/4, -7/8)# Explanation: # y= 2x^2-x-3-4(x-1)^2# #y = 2x^2-x-3-4x^2+8x-4# #y = -2x^2 + 7x - 7# #f(x) = ax^2 + bx + c " : x vertex" = (-b)/(2a)# #(-b)/(2a) = (-7)/(2(-2) = 7/4# #y = -2(7/4)^2 + 7(7/4) - 7 = (-7)/8# 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 1405 views around the world You can reuse this answer Creative Commons License