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What is the largest value of a for which we can prove that the inequality below is true for all values of x?

x²+8x+21≥ a

User Mikku
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Final answer:

The largest value of 'a' that makes the inequality x² + 8x + 21 ≥ a true for all values of x is 5, which is found by completing the square and locating the vertex of the parabola.

Step-by-step explanation:

The largest value of 'a' for which the inequality x² + 8x + 21 ≥ a is true for all values of x can be found by completing the square or by looking at the vertex of the parabola represented by the quadratic equation. This quadratic represents a parabola that opens upwards (since the coefficient of x² is positive), and its minimum value occurs at the vertex. To find the vertex, we complete the square:

x² + 8x + 21 = (x + 4)² - 16 + 21

(x + 4)² + 5

The vertex form of the parabola is (x + 4)² + 5, which means its vertex is at (-4, 5), and the minimum value of this parabola is 5. Therefore, the inequality holds for all x if and only if a ≤ 5. The largest value of 'a' that satisfies this condition is a = 5.

User Aaron Bonner
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