Find the positive and negative domains of the function below:
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Find the positive and negative domains of the function below:
To solve this problem, follow these steps:
Now, work through these steps:
Step 1: Recognize the quadratic function as a perfect square trinomial. It can be rewritten as .
Step 2: The vertex of the parabola, which also represents its minimum point since the parabola opens upwards, occurs at .
Step 3: Since for all real (because a square is always non-negative), the function is non-negative everywhere.
Therefore, the function is never negative, and since it equals zero at , it is positive for all .
Therefore, the function's positive domain is . For , the function is not negative, hence there is no such domain.
The solution to the problem is:
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The graph of the function below does not intersect the \( x \)-axis.
The parabola's vertex is marked A.
Find all values of \( x \) where
\( f\left(x\right) > 0 \).
The positive domain is all x-values where y is positive (y > 0), not where x > 0! Since is positive everywhere except at x = -5, the positive domain includes both positive and negative x-values.
At x = -5, we get . Since 0 is neither positive nor negative, x = -5 belongs to neither the positive nor negative domain.
Since is a perfect square, it's always non-negative (≥ 0). Perfect squares can equal zero but never go below zero, so this function has no negative values.
When y = 0, the function touches the x-axis but isn't negative. The negative domain would be where y < 0 (below the x-axis). Since this parabola only touches but never goes below the x-axis, there's no negative domain.
No! While the function is non-negative everywhere, it equals exactly zero at x = -5. The positive domain only includes where y > 0, so we must exclude the point where y = 0.
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