What is the vertex form of #y=1/5x^2-4/7x+3/5#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer jos Jul 11, 2018 #y=1/5(x-10/7)^2+47/245# Explanation: #1/5x^2-4/7x+3/5# #=1/5(x^2-20/7x)+3/5# #=1/5(x^2-20/7x+(20/7divide2)^2-(20/7divide2)^2)+3/5# #=1/5(x-10/7)^2-1/5*100/49+3/5# #=1/5(x-10/7)^2+(3*49-100)/(5*49)# #=1/5(x-10/7)^2+47/245# 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 1551 views around the world You can reuse this answer Creative Commons License