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Find f(x)=x²- 6x+9 where f(x)>0 and f(x)<0?

User Quamrana
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Final answer:

The function f(x) = x² - 6x + 9 is always greater than or equal to zero because it is a perfect square trinomial with a double root at x = 3. Therefore, f(x) > 0 for all x except at x = 3, where f(x) = 0. There are no values of x for which f(x) < 0.

Step-by-step explanation:

To find where the function f(x) = x² - 6x + 9 is greater than zero (f(x) > 0) and less than zero (f(x) < 0), we need to analyze the roots of the equation and the nature of the quadratic function. First, we factor the quadratic equation:

f(x) = (x - 3)²

This is a perfect square trinomial, and it is clear that the function has a double root at x = 3. Therefore, f(x) is always non-negative, since a square of a real number is non-negative. This implies that f(x) ≥ 0 for all x and f(x) is never less than zero, hence there are no values of x for which f(x) < 0.

However, we also know that f(x) is exactly zero when x = 3, which is at the vertex of the parabola. The graph of this function is a parabola opening upwards, and since there are no x-values where the function dips below the x-axis, f(x) > 0 for all x except for x = 3.

User Carl Groner
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