What is the vertex of y= 1/3(x/5+1)^2+4/15 ? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs 1 Answer Karthik G Jan 2, 2016 The vertex form is y = a(x-h)^2 +k where (h,k) is the vertex. For our problem the vertex is (-5,4/15) Explanation: y=1/3(x/5+1)^2 + 4/15 y=1/3((x+5)/5)^2 + 4/15 y=1/75(x+5)^2 + 4/15 Compare with y=a(x-h)^2 + k h=-5 and k=4/15 The vertex (h,k) is (-5,4/15) 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 1578 views around the world You can reuse this answer Creative Commons License