Calculate Negative Area: Finding Area Below f(x) = x² + 16

Q: What does 'negative area' actually mean?

Negative area refers to regions where the function is below the x-axis (where f(x)

Q: Why is this function always positive?

The function $ f(x) = x^2 + 16 $ has two parts: $ x^2 $ (always ≥ 0) and +16. Even when $ x^2 = 0 $, we still have f(x) = 16, which is positive!

Q: How do I find the vertex of any parabola?

For $ f(x) = ax^2 + bx + c $, the vertex x-coordinate is $ x = -\frac{b}{2a} $. Here, a=1 and b=0, so $ x = -\frac{0}{2(1)} = 0 $.

Q: Could this parabola ever have negative area?

No, never! Since the constant term is +16 and the coefficient of $ x^2 $ is positive, this parabola opens upward and has its minimum at y = 16. It never touches the x-axis.

Q: What if the question asked about a different function like x² - 16?

$ f(x) = x^2 - 16 $ would have negative area!It equals zero when x = ±4Between x = -4 and x = 4, the function is negative

Quadratic Functions with Always-Positive Behavior

Find the negative area of the function

$f(x)=x^2+16$

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:05 Let's find where the function is below the X-axis.

00:09 Notice the X squared term is positive. So, the function smiles.

00:14 The negative domain is where the graph dips under the X-axis.

00:20 To find this, set Y to zero. Let's find the X-intercepts.

00:25 Next, we isolate X. Keep watching.

00:29 Remember, any number squared is always positive.

00:35 Therefore, no X-intersects exist for our graph.

00:41 Our function remains entirely above the X-axis.

00:45 This means the domain is always positive.

00:49 The function is always above the X-axis, so it's always positive.

00:57 That's how we solve this problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the negative area of the function

$f(x)=x^2+16$

Step-by-step solution

To solve this problem, we'll follow the steps outlined in our analysis.

Step 1: Analyze the function's form $f(x) = x^2 + 16$ . Here, $a=1, b=0,$ and $c=16$ .
Step 2: Find the vertex to see if the function ever takes negative values. The vertex $x$ is calculated by $x = -\frac{b}{2a} = 0$ .
Step 3: Evaluate $f(x)$ at this vertex: $f(0) = 0^2 + 16 = 16$ .
Step 4: Determine when $f(x) < 0$ . Since $f(x) = x^2 + 16 \geq 16$ for all real numbers $x$ , the function is always positive.
Step 5: Compare the finding against multiple-choice options. The choice indicating that $f(x)$ is always positive is the correct one: "Always positive".

The conclusion, therefore, is as follows: the function $f(x) = x^2 + 16$ is always positive, and there is no negative area under the graph relative to the x-axis.

Final Answer

Always positive

Key Points to Remember

Essential concepts to master this topic

Vertex Rule: Find minimum using $x = -\frac{b}{2a}$
Technique: Evaluate at vertex: $f(0) = 0^2 + 16 = 16$
Check: Since $x^2 \geq 0$ always, $f(x) = x^2 + 16 \geq 16$ for all x ✓

Common Mistakes

Avoid these frequent errors

Confusing negative area with function values below zero
Don't think negative area means finding where f(x) < 0 on this upward parabola = impossible search! This function never goes below the x-axis because its minimum value is 16. Always check if the parabola actually dips below zero before looking for negative regions.

Practice Quiz

Test your knowledge with interactive questions

Which chart represents the function $$ y=x^2-9 $$ ?

FAQ

Everything you need to know about this question

What does 'negative area' actually mean?

Negative area refers to regions where the function is below the x-axis (where f(x) < 0). Since this parabola stays above the x-axis, there is no negative area to calculate.

Why is this function always positive?

The function $f(x) = x^2 + 16$ has two parts: $x^2$ (always ≥ 0) and +16. Even when $x^2 = 0$ , we still have f(x) = 16, which is positive!

How do I find the vertex of any parabola?

For $f(x) = ax^2 + bx + c$ , the vertex x-coordinate is $x = -\frac{b}{2a}$ . Here, a=1 and b=0, so $x = -\frac{0}{2(1)} = 0$ .

Could this parabola ever have negative area?

No, never! Since the constant term is +16 and the coefficient of $x^2$ is positive, this parabola opens upward and has its minimum at y = 16. It never touches the x-axis.

What if the question asked about a different function like x² - 16?

$f(x) = x^2 - 16$ would have negative area!
It equals zero when x = ±4
Between x = -4 and x = 4, the function is negative

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Parabola Families questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime