Answer:
D. Minimum at (3, 7)
Step-by-step explanation:
We can add and subtract the square of half the x-coefficient:
y = x^2 -6x +(-6/2)^2 +16 -(-6/2)^2
y = (x -3)^2 +7 . . . . . simplify to vertex form
Comparing this to the vertex for for vertex (h, k) ...
y = (x -h)^2 +k
We find the vertex to be ...
(3, 7) . . . . vertex
The coefficient of x^2 is positive (+1), so the parabola opens upward and the vertex is a minimum.
Answer:
Step-by-step explanation:
If the roots are 1 + 5i and 1 - 5i, then you need the factors that result from those roots. They are (x - 1 + 5i) and (x - 1 - 5i). Now what you do with those is FOIL them out. Doing that gives you the following:
(what a mess, huh?)
The good thing is that several of those terms cancel each other out. +5ix cancels out the -5ix; -5i cancels out the 5i; and the 2 -x terms combine to -2x. That leaves you with:
Obviously you're in the section in math that deals with complex (imaginary) numbers so you should know that i-squared is equal to -1. Making that replacement:
a = 1, b = -2, c = 25