What is the vertex form of # y= (6x-6)(x+2)+4x^2+5x#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer Binayaka C. Nov 23, 2017 Vertex form of equation is #y= 10(x+0.55)^2-15.025 # Explanation: #y=(6x-6)(x+2)+4x^2+5x # or #y=6x^2+12x-6x-12+4x^2+5x # or #y=10x^2+11x-12 or y= 10(x^2+11/10x)-12 # or #y= 10{x^2+11/10x+(11/20)^2}-10*(11/20)^2-12 # or #y= 10(x+11/20)^2-3.025-12 # or #y= 10(x+0.55)^2-15.025 # .Comparing with standard vertex form of equation # f(x)=a(x-h)^2+k ; (h,k)# being vertex we find here #h= -0.55 , k=-15.025 # So vertex is at #(-0.55,-15.025)# and vertex form of equation is #y= 10(x+0.55)^2-15.025 # [Ans] 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 1183 views around the world You can reuse this answer Creative Commons License