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 1656 views around the world You can reuse this answer Creative Commons License