How do you find the minimum and maximum value of #y=-(x-1)(x+4)#? Algebra Quadratic Equations and Functions Vertex Form of a Quadratic Equation 1 Answer Ultrilliam Jul 28, 2018 #y=-(x-1)(x+4) = -(x^2 + 3x - 4)# Completing the square: # = -((x + 3/2)^2 - 9 /4- 4)# # = -((x + 3/2)^2 - 25 /4 )# #implies y = -(x + 3/2)^2 + 25 /4 # Now: #(x + 3/2)^2 ge 0 qquad forall x# #:. - (x + 3/2)^2 le 0 qquad forall x# #:. y le 25/4 qquad forall x# #implies {(y_("max") = 25/4),(y_("max") = - oo):}# graph{-(x-1)(x+4) [-5, 5, -9.01, 9.01]} 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 1991 views around the world You can reuse this answer Creative Commons License