How do you find the maximum and minimum of #y=(x+2)^2-3#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer Binayaka C. Dec 21, 2016 Maximum: #oo# , Minimum: # -3# Explanation: #y=(x+2)^2-3 = x^2+4x+1# Comparing with quadratic equation #y= ax^2+bx+c # Here #a=1 , b= 4 ,c =1 #.Discriminant #D= b^2-4ac=4^2-4*1*1 =12 # If #a>0# ; maximum is #oo# and minimum is #y=-D/(4a)=-12/4=-3# graph{x^2+4x+1 [-10, 10, -5, 5]}[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 3568 views around the world You can reuse this answer Creative Commons License