2006-10-28 02:01:07 · 4 answers · asked by MANDY 1 in Science & Mathematics ➔ Mathematics
easy
you mean x^2-6x -13 (no zero at end)
= (x-3)^2-22
so lowest value is 22 as (x-3)^2 lowest is 0 asuming x real
if x is not real any value is possible
2006-10-28 02:07:50 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
The first derivative of the equation x^2 - 6x - 13 is 2x - 6
Solving 2x-6 for its absolute minimum, solve the equation 2x-6=0.
You get x = 3.
Plug 3 into the original equation: (3)^2 - 6(3) - 13 = 9 - 18 - 13
and 9-18-13 = -22
So the answer is -22
2006-10-28 09:22:02 · answer #2 · answered by Dan 2 · 0⤊ 0⤋
x^2 - 6x -13 = x^2 - 6x + 9 - 22 = (x-3)^2 - 22 >= -22
the least value is -22 when x = 3
2006-10-28 09:30:45 · answer #3 · answered by James Chan 4 · 0⤊ 0⤋
x=(-6-4i)/2
=2(-3-2i)/2
2006-10-28 09:05:57 · answer #4 · answered by Anonymous · 0⤊ 0⤋