Calculate Negative Area of f(x) = x² - 16: Below X-Axis Integration

Q: How do I know which interval makes the function negative?

After finding the zeros (-4 and 4), test a point in each interval. Pick x = 0 (between -4 and 4): $ f(0) = 0^2 - 16 = -16 $. Since this is negative, the function is negative on \( -4 .

Q: Why does the parabola go below the x-axis between the roots?

Since the coefficient of $ x^2 $ is positive (1), this parabola opens upward. It starts high, crosses the x-axis at x = -4, dips below (negative area), then crosses back up at x = 4.

Q: What if I need to find the actual negative area value?

To find the area, you'd integrate: $ -\int_{-4}^{4} (x^2 - 16) dx $. The negative sign ensures you get a positive area value since the function is negative in this interval.

Q: Can I solve this without factoring?

Yes! Use the quadratic formula: $ x = \frac{0 \pm \sqrt{0 + 64}}{2} = \pm 4 $. But factoring $ x^2 - 16 = (x-4)(x+4) $ is usually faster for perfect squares.

Q: What does 'negative area' actually mean?

Negative area refers to the region where the function values are negative (below the x-axis). It's the interval where \( f(x) , not a calculation of actual area.

Quadratic Functions with Negative Region Analysis

Find the negative area of the function

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

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

00:09 The X squared coefficient is positive. So the curve is happy and opens upward.

00:15 Remember, the negative domain is the part under the X-axis.

00:20 To find X-intercepts, set Y equal to zero.

00:24 Now, let's isolate X in the equation.

00:28 Next step, take the square root. Remember, it gives both a positive and negative solution.

00:36 These solutions are where the graph crosses the X-axis.

00:42 Let's put these crossing points on the graph.

00:48 We can see the positive domain is above the X-axis.

00:59 And remember, the negative domain is what lies below the X-axis.

01:04 Let's focus on finding the negative domain, where the graph dips below.

01:10 And that's how we find where the function is negative!

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 find where the function $f(x) = x^2 - 16$ is negative.

Step 1: Set $f(x) = 0$ to find where the function has zero values, solving $x^2 - 16 = 0$ .
Step 2: Factor the equation as $(x - 4)(x + 4) = 0$ .
Step 3: Find the roots: $x = 4$ and $x = -4$ .
Step 4: Test intervals: $x < -4$ , $-4 < x < 4$ , and $x > 4$ . Inside the interval $-4 < x < 4$ , $x^2 - 16$ is negative because the square of any number less than 4 but greater than -4 will be less than 16.

Therefore, the function $f(x)$ is negative on the interval $-4 < x < 4$ .

Final Answer

$-4 < x < 4$

Key Points to Remember

Essential concepts to master this topic

Rule: Function is negative where graph lies below x-axis
Technique: Find zeros by factoring: $x^2 - 16 = (x-4)(x+4) = 0$
Check: Test value x = 0: $0^2 - 16 = -16 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Confusing where function is negative vs positive
Don't assume the function is negative where x is negative = wrong interval! The sign of x doesn't determine the sign of f(x). Always test intervals between zeros to find where the parabola dips below the x-axis.

Practice Quiz

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One function

$$ y=-6x^2 $$

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FAQ

Everything you need to know about this question

How do I know which interval makes the function negative?

After finding the zeros (-4 and 4), test a point in each interval. Pick x = 0 (between -4 and 4): $f(0) = 0^2 - 16 = -16$ . Since this is negative, the function is negative on $-4 < x < 4$ .

Why does the parabola go below the x-axis between the roots?

Since the coefficient of $x^2$ is positive (1), this parabola opens upward. It starts high, crosses the x-axis at x = -4, dips below (negative area), then crosses back up at x = 4.

What if I need to find the actual negative area value?

To find the area, you'd integrate: $-\int_{-4}^{4} (x^2 - 16) dx$ . The negative sign ensures you get a positive area value since the function is negative in this interval.

Can I solve this without factoring?

Yes! Use the quadratic formula: $x = \frac{0 \pm \sqrt{0 + 64}}{2} = \pm 4$ . But factoring $x^2 - 16 = (x-4)(x+4)$ is usually faster for perfect squares.

What does 'negative area' actually mean?

Negative area refers to the region where the function values are negative (below the x-axis). It's the interval where $f(x) < 0$ , not a calculation of actual area.

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