Determine x for the Quadratic Function y = x² + 16 When f(x) > 0

Q: Why doesn't this quadratic have any zeros like other quadratics?

This quadratic never crosses the x-axis because it's always above it! The constant term (+16) shifts the basic parabola $ y = x^2 $ up by 16 units, so it never touches zero.

Q: How do I know the minimum value without graphing?

Since $ x^2 $ is always non-negative, its smallest value is 0 (when x = 0). So the minimum of $ x^2 + 16 $ is 0 + 16 = 16.

Q: What if the question asked when f(x) < 0?

The answer would be "No x" or "Never" because this function is always positive. It never goes below the x-axis!

Q: Are there quadratics that are positive for only some x-values?

Yes! Quadratics like $ x^2 - 4 $ are positive when $ x > 2 $ or \( x , and negative between the roots. The key is whether the constant term makes the minimum positive or negative.

Q: How can I verify that ALL x-values work?

Test several different values: $ x = 0 $ gives 16, $ x = 3 $ gives 25, $ x = -10 $ gives 116. Since all results are positive, the pattern holds for all real numbers.

Quadratic Functions with Always Positive Values

Given the function:

$y=x^2+16$

Determine for which values of x is $f(x) > 0$ true

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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 is $f(x) > 0$ true

Step-by-step solution

To solve the problem of determining for which values of $x$ the function $y = x^2 + 16$ is positive, we proceed as follows:

Step 1: Analyze the function $y = x^2 + 16$ .
The expression $x^2$ is non-negative (i.e., $x^2 \geq 0$ ) for all real numbers $x$ . Therefore, the smallest value $x^2$ can take is 0.

Step 2: Evaluate the function at its minimum value.
Substituting the minimum value of $x^2$ into the function gives us:

$y = x^2 + 16 = 0 + 16 = 16$ .

Step 3: Determine for which $x$ the function is positive.
Since the minimum value of the function is 16, which is greater than 0, the function $y = x^2 + 16$ is greater than 0 for all real numbers $x$ .

Thus, the solution to the problem is that the function is positive for all $x$ .

Final Answer

All x

Key Points to Remember

Essential concepts to master this topic

Rule: Quadratic functions with positive discriminants have minimum values
Technique: For $x^2 + 16$ , minimum occurs when $x^2 = 0$ , giving $y = 16$
Check: Test any x-value: $(-5)^2 + 16 = 25 + 16 = 41 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Setting the function equal to zero to find when it's positive
Don't solve $x^2 + 16 = 0$ to find when f(x) > 0! This equation has no real solutions, leading students to think the function is never positive. Always find the minimum value by recognizing that $x^2 \geq 0$ , so the smallest possible value is 16.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of $$ x $$ where $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why doesn't this quadratic have any zeros like other quadratics?

This quadratic never crosses the x-axis because it's always above it! The constant term (+16) shifts the basic parabola $y = x^2$ up by 16 units, so it never touches zero.

How do I know the minimum value without graphing?

Since $x^2$ is always non-negative, its smallest value is 0 (when x = 0). So the minimum of $x^2 + 16$ is 0 + 16 = 16.

What if the question asked when f(x) < 0?

The answer would be "No x" or "Never" because this function is always positive. It never goes below the x-axis!

Are there quadratics that are positive for only some x-values?

Yes! Quadratics like $x^2 - 4$ are positive when $x > 2$ or $x < -2$ , and negative between the roots. The key is whether the constant term makes the minimum positive or negative.

How can I verify that ALL x-values work?

Test several different values: $x = 0$ gives 16, $x = 3$ gives 25, $x = -10$ gives 116. Since all results are positive, the pattern holds for all real numbers.

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