Find the Area of y=(x+5)²-4: Quadratic Function Analysis

Q: Why is the function positive outside the roots instead of between them?

Since this is a parabola opening upward (positive coefficient of x²), it's shaped like a U. The function dips below the x-axis between the roots (-7 and -3) and rises above it outside this interval.

Q: How do I know which intervals to test?

The roots divide the number line into three parts: everything left of -7, between -7 and -3, and right of -3. Pick any test point from each interval to see if the function is positive or negative there.

Q: What does 'positive area' actually mean?

Positive area means the regions where the function value y is greater than zero. You're finding where the graph sits above the x-axis, not calculating actual area under a curve.

Q: Can I solve this by expanding the squared term?

Yes! Expanding gives $ y = x^2 + 10x + 21 $. Set this > 0 and factor to get the same roots. However, vertex form is often faster for identifying the transformation.

Q: Why do we use 'or' instead of 'and' in the final answer?

The function is positive in two separate regions: x -3. Using 'and' would mean both conditions must be true simultaneously, which is impossible since no number can be both less than -7 and greater than -3.

Quadratic Inequalities with Positive Area

Find the positive area of the function

$y=(x+5)^2-4$

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Understand the problem

Find the positive area of the function

$y=(x+5)^2-4$

Step-by-step solution

To find the positive area of the function $y = (x+5)^2-4$ , follow these steps:

Step 1: Identify when the function is greater than zero.
We need $(x+5)^2 - 4 > 0$ .
Step 2: Find the roots of $(x+5)^2 - 4 = 0$ .
Solving, we set: $(x+5)^2 = 4$ .
Step 3: Solve for $(x+5)^2 = 4$ .
Take the square root on both sides: $x + 5 = \pm 2$ .
This gives: $x + 5 = 2$ or $x + 5 = -2$ .
Thus, $x = -3$ or $x = -7$ .
Step 4: Identify intervals.
We need to look at intervals determined by the roots: $(-\infty, -7)$ , $(-7, -3)$ , and $(-3, \infty)$ .
Step 5: Determine where the function is positive by testing each interval:
- For $x < -7$ , choose $x = -8$ , then $((-8)+5)^2 - 4 = 9 - 4 = 5$ (positive).
- For $-7 < x < -3$ , choose $x = -5$ , then $((-5)+5)^2 - 4 = 0 - 4 = -4$ (negative).
- For $x > -3$ , choose $x = 0$ , then $(5)^2 - 4 = 25 - 4 = 21$ (positive).

Therefore, the positive area of the function is for $x < -7$ and $-3 < x$ .

Final Answer

$x < -7 , -3 < x$

Key Points to Remember

Essential concepts to master this topic

Rule: Find where quadratic function is positive by solving inequality
Technique: Set $(x+5)^2 - 4 > 0$ and find roots x = -7, -3
Check: Test values: x = -8 gives 5 > 0, x = 0 gives 21 > 0 ✓

Common Mistakes

Avoid these frequent errors

Writing interval notation incorrectly
Don't write -3 < x < -7 when the function is positive outside the roots = impossible interval! This creates a contradiction since -3 cannot be less than -7. Always identify that the parabola opens upward, so it's positive outside the roots: x < -7 or x > -3.

Practice Quiz

Test your knowledge with interactive questions

Find the corresponding algebraic representation of the drawing:

FAQ

Everything you need to know about this question

Why is the function positive outside the roots instead of between them?

Since this is a parabola opening upward (positive coefficient of x²), it's shaped like a U. The function dips below the x-axis between the roots (-7 and -3) and rises above it outside this interval.

How do I know which intervals to test?

The roots divide the number line into three parts: everything left of -7, between -7 and -3, and right of -3. Pick any test point from each interval to see if the function is positive or negative there.

What does 'positive area' actually mean?

Positive area means the regions where the function value y is greater than zero. You're finding where the graph sits above the x-axis, not calculating actual area under a curve.

Can I solve this by expanding the squared term?

Yes! Expanding gives $y = x^2 + 10x + 21$ . Set this > 0 and factor to get the same roots. However, vertex form is often faster for identifying the transformation.

Why do we use 'or' instead of 'and' in the final answer?

The function is positive in two separate regions: x < -7 OR x > -3. Using 'and' would mean both conditions must be true simultaneously, which is impossible since no number can be both less than -7 and greater than -3.

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