Question

(a) Express f(x) = x^2 - 8x + 22 in the form f(x) = (x +a)^2 +b, where a and b are constants to be determined. (b) State the minimum value of f(x) and the value of x at which the minimum occurs. (c) Solve the inequality x^2 - 8x + 22 > 10.

Answer 1

• (x -4)² +6
• minimum value: at x = 4
• (x < 2) ∪ (6 < x) . . . for f(x) > 10

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

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(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.

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(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).

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You can see the vertex in the attached graph, and that f(x) > 10 for x outside the interval [2, 6].