What is the the vertex of #y =-x^2-12x-14 #?

1 Answer
Jan 3, 2016

Complete the square to find the vertex.

Explanation:

y = -#x^2# - 12x - 14
y = -(#x^2# + 12x) - 14
y = -(#x^2# + 12x + n - n) - 14
n = #(b/2)^2#
n = #(12/2)^2#
n = 36
y = -(#x^2# + 12x + 36 - 36) - 14
y = -(#x^2# + 12x + 36) + 36 - 14
y = -#(x + 6)^2# + 22

In the form y = a#(x - p)^2# + q, the vertex is at (p, q)

So, the vertex of #-x^2# - 12x - 14 is (-6, 22).

Hopefully you understand now