How do you find the vertex of #y = x^2 - 2#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer Konstantinos Michailidis Mar 3, 2016 The vertex form is #y=a*(x-h)^2+k# hence for #y=x^2-2# is #a=1 , h=0, k=-2# So the vertex is #(h,k)=(0,-2)# 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 4206 views around the world You can reuse this answer Creative Commons License