How do you write the quadratic in vertex form given #f(x)=x^2 + 6x + 12#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer George C. May 12, 2015 #f(x) = x^2 + 6x + 12# #= (x+3)^2 - 9 + 12# #= (x+3)^2 + 3# In general, #ax^2 + bx+ c# can be written in vertex form as #a(x+(b/(2a)))^2+(c-b^2/(4a))# In the case #a=1#, this simplifies to #x^2+bx+c = (x+b/2)^2 + (c - b^2/4)# 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 1698 views around the world You can reuse this answer Creative Commons License