Identify Negative Values for the Function y = x² + 16: An Inequality Challenge

Q: Why can't x² + 16 ever be negative?

Because x² is always ≥ 0 for any real number! When you add 16 to something that's already non-negative, the result must be at least 16.

Q: What if I try x = -4 or x = 4?

Let's check: $ (-4)^2 + 16 = 16 + 16 = 32 $ and $ 4^2 + 16 = 16 + 16 = 32 $. Both are positive, not negative!

Q: How do I know when a quadratic has no negative values?

Look for the form $ ax^2 + c $ where a > 0 and c > 0. Since x² ≥ 0, adding a positive constant keeps everything positive!

Q: Is this different from finding where the graph is below the x-axis?

That's exactly what we're looking for! The graph of $ y = x^2 + 16 $ is a parabola opening upward with its lowest point at (0, 16), so it never touches or goes below the x-axis.

Quadratic Functions with Always-Positive Values

Given the function:

$y=x^2+16$

Determine for which values of x $f(x) < 0$ holds

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function:

$y=x^2+16$

Determine for which values of x $f(x) < 0$ holds

Step-by-step solution

We start by considering the function given: $y = x^2 + 16$ . Our task is to find the values of $x$ making $f(x) < 0$ , i.e., $x^2 + 16 < 0$ .

Let's analyze the expression:

The square of any real number $x$ , denoted as $x^2$ , is always non-negative. Thus, $x^2 \geq 0$ for all $x \in \mathbb{R}$ .
Adding 16 to a non-negative number (like $x^2$ ) results in a number that is at least 16, i.e., $x^2 + 16 \geq 16$ .

Thus, the expression $x^2 + 16$ is always at least 16, and there are no $x$ values for which $y < 0$ .

The solution is that there are No x values where $f(x) < 0$ .

Hence, option 4 is correct: No x.

Final Answer

No x

Key Points to Remember

Essential concepts to master this topic

Rule: x² is always non-negative for all real numbers
Technique: Add positive constant: x² + 16 ≥ 16 for all x
Check: Test any value like x = 0: 0² + 16 = 16 > 0 ✓

Common Mistakes

Avoid these frequent errors

Solving x² + 16 = 0 to find negative values
Don't try to solve x² + 16 = 0 by subtracting 16 = getting x² = -16 with no real solutions! This misses the point that we need f(x) < 0, not f(x) = 0. Always recognize that x² + positive number is always positive.

Practice Quiz

Test your knowledge with interactive questions

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
$f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why can't x² + 16 ever be negative?

Because x² is always ≥ 0 for any real number! When you add 16 to something that's already non-negative, the result must be at least 16.

What if I try x = -4 or x = 4?

Let's check: $(-4)^2 + 16 = 16 + 16 = 32$ and $4^2 + 16 = 16 + 16 = 32$ . Both are positive, not negative!

Could there be complex solutions?

The question asks for real values of x where the function is negative. Even with complex numbers, we're looking at a real-valued function that's always positive for real inputs.

How do I know when a quadratic has no negative values?

Look for the form $ax^2 + c$ where a > 0 and c > 0. Since x² ≥ 0, adding a positive constant keeps everything positive!

Is this different from finding where the graph is below the x-axis?

That's exactly what we're looking for! The graph of $y = x^2 + 16$ is a parabola opening upward with its lowest point at (0, 16), so it never touches or goes below the x-axis.

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