Question

express f(x)= -x^2 - 12x + 22 in the form a(x - h)2 + k

Answer 1

Hi, Lauren,
Have you used "completing the square" before when addressing a quadratic expression?

Rewrite f(x)= -x^2 - 12x + 22 as -1(x^2 + 12x ) +22
Reasons: To identify a in the form a(x-h)^2 + k, and to be able to focus on the relatively simple quadratic x^2 + 12x.

To "complete the square" of x^2 + 12x, take HALF of the coefficient of x; that is, take half of 12 (this results in 6). Now square that 6 (which results in 36) and add that to x^2 + 12x , and then subtract 36:

f(x)= -x^2 - 12x + 22 = -1(x^2 +6x +36 -36) +22
Now, x^2 + 6x + 36 can be written as (x+6)^2, and so
f(x)= -x^2 - 12x + 36 becomes -( (x+6)^2 -36 ) + 22.

Answer 2

Simplifying, we get f(x) -(x+6)^2 +36 + 22, or -(x+6)^2 + 58.

Answer 3

To determine the coefficients of the vertex, re-write this f(x) as
f(x) = -(x-[-6])^2 +36 + 22, or f(x) = -(x-[-6])^2 +58. This has the desired form a(x - h)2 + k.
Comparing these two expressions, we see that a = -1; h = -2, and k = 58.
The vertex of this parabola is at (-6,58). Because a is (-), the graph of this parabola opens downward.