What is the vertex of # y= -3x^2 - 2x - (x+2)^2 #?

Question

What is the vertex of #  y= -3x^2 - 2x - (x+2)^2 #?

Eva Adamson · Accepted Answer

The vertex is at #(-3/4,-7/4)# #y=-3x^2-2x-(x+2)^2# Expand the polynomial: #y=-3x^2-2x-(x^2+4x+4)# Combine like terms: #y=-4x^2-6x-4# Factor out #-4#: #y=-4[x^2+3/2x+1]# Complete the square: #y=-4[(x+3/4)^2-(3/4)^2+1]# #y=-4[(x+3/4)^2+7/16]# #y=-4(x+3/4)^2-7/4# From vertex form, the vertex is at #(-3/4,-7/4)#

Eva Adamson · Answer

Vertex: #(-3/4, -55/16)~~(-0.75, -3.4375)#

1) Rewrite this equation in standard form #y=-3x^2-2x-(x+2)^2# #y=-3x^2-2x-(x^2+4x+4)# #y=-4x^2-6x-4#

2) Rewrite this equation in vertex form by completing the square #y=(-4x^2-6x)-4# #y=-4(x^2+3/2x)-4# #y=-4(x^2+3/2x+(3/4)^2)-4+(3/4)^2# #y=-4(x+3/4)^2-55/16#

The vertex form is #y=a(x-h)^2+k# reveals the vertex at #(h,k)#

Vertex: #(-3/4, -55/16)~~(-0.75, -3.4375)#

You can see this if you graph the equation graph{y=-4x^2-6x-4 [-3, 2, -7, 0.1]}

Nathan Axel · Answer

undefined To find the vertex of the quadratic function $y = -3x^2 - 2x - (x+2)^2$, first, rewrite the expression in standard form: $y = -4x^2 - 4x - 4$. The vertex of a quadratic function in the form $y = ax^2 + bx + c$ is given by the coordinates $(h, k)$, where $h = -\frac{b}{2a}$ and $k = f(h)$. Substitute $h$ into the function to find $k$. Thus, the vertex of the function is $(-0.5, -3)$.