What is the the vertex of #x =-1/2(y-2)^2-4 #? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs 1 Answer Tony B Apr 11, 2016 vertex#=(-4,2)# Explanation: #x=-1/2(ycolor(green)(-2))^2color(red)(-4)# Consider the #color(green)( 2)# from #(ycolor(green)(-2))# #y_("vertex")=(-1)xxcolor(green)(-2)=+2# #x_("vertex")=color(red)(-4)# 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 1084 views around the world You can reuse this answer Creative Commons License