Answer:
- (x -4)² +6
- minimum value: at x = 4
- (x < 2) ∪ (6 < x) . . . for f(x) > 10
Step-by-step explanation:
(a) You can split the constant so it includes a portion that is the square of half the x-coefficient:
f(x) = x² -8x +16 +6
Now, that first trinomial can be written as the square you want:
f(x) = (x -4)² +6
__
(b) The form in part (a) is "vertex form" where (-a, b) is the vertex. When the coefficient of x² is positive, as here, the parabola opens upward, and its vertex is its minimum.
The minimum occurs at (4, 6), where x=4.
__
(c) Using the above rewrite of f(x), we have the inequality ...
f(x) > 10
(x -4)² +6 > 10 . . . substitute the vertex form of f(x)
(x -4)² > 4 . . . . . . . subtract 6
|x -4| > 2 . . . . . . . . take the square root
This means two things:
-2 > x -4 ⇒ 2 > x
and
x -4 > 2 ⇒ x > 6
Then the solution to the inequality is (x < 2) ∪ (6 < x).
_____
You can see the vertex in the attached graph, and that f(x) > 10 for x outside the interval [2, 6].
