What is the vertex of # y=1-4x-x^2 #? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer ali ergin Jul 21, 2016 {-2,5} Explanation: #y=1-4x-x^2# #(d y)/(d x)=0-4-2x=0# #-4-2x=0# #2x=-4" ; "x=-4/2=-2# #y=1-4(-2)-(-2)^2# #y=1+8-4#=5 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 2978 views around the world You can reuse this answer Creative Commons License