Finding Negative Values of y: Solve for x in y = -x² + 49

Q: How do I know which intervals make the function negative?

Since $ y = -x^2 + 49 $ is a downward-opening parabola (negative coefficient of x²), it's positive between the roots and negative outside them. Test a point in each region to confirm!

Q: Why do we solve -x² + 49 = 0 first?

The x-intercepts are where the function changes from positive to negative (or vice versa). These boundary points at x = -7 and x = 7 divide the number line into regions we need to test.

Q: What's the difference between < and ≤ in the final answer?

Since we want f(x) strict inequality signs. The points x = -7 and x = 7 make f(x) = 0, so they're not included in our solution.

Q: How can I visualize this problem?

Picture an upside-down parabola with vertex at (0, 49) crossing the x-axis at x = -7 and x = 7. The function is negative where the graph dips below the x-axis - that's outside the interval!

Q: What if I get confused about which way the parabola opens?

Look at the coefficient of x²! If it's negative (like -1 here), the parabola opens downward. If it's positive, it opens upward. This determines where the function is positive vs negative.

Quadratic Inequalities with Sign Analysis

Look at the following function:

$y=-x^2+49$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$y=-x^2+49$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

Step-by-step solution

The solution to the problem involves finding the values of $x$ where the function $y = -x^2 + 49$ is less than zero. Since it is a downward-opening parabola, its intercepts tell us where the function changes sign.

To start, solve for $f(x) = 0$ :

$-x^2 + 49 = 0$

Add $x^2$ to both sides:

$x^2 = 49$

Take the square root of both sides:

$x = \pm 7$

These solutions $x = 7$ and $x = -7$ are the x-intercepts of the parabola. Because the parabola opens downwards, the function is negative outside this interval.

Thus, the function $f(x) < 0$ for the intervals:

$x > 7$
$x < -7$

Therefore, the solution to the problem is:

$x > 7$ or $x < -7$

Final Answer

$x > 7$ or $x < -7$

Key Points to Remember

Essential concepts to master this topic

Rule: Find x-intercepts by setting the quadratic equal to zero
Technique: Test signs between intervals: -x² + 49 < 0 outside roots
Check: Test x = 8: -(8)² + 49 = -15 < 0 ✓

Common Mistakes

Avoid these frequent errors

Misinterpreting the parabola's orientation and intervals
Don't assume f(x) < 0 means the middle interval between roots! Since y = -x² + 49 opens downward, it's negative outside the interval [-7, 7], not inside. Always consider whether the parabola opens up or down when determining sign intervals.

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

How do I know which intervals make the function negative?

Since $y = -x^2 + 49$ is a downward-opening parabola (negative coefficient of x²), it's positive between the roots and negative outside them. Test a point in each region to confirm!

Why do we solve -x² + 49 = 0 first?

The x-intercepts are where the function changes from positive to negative (or vice versa). These boundary points at x = -7 and x = 7 divide the number line into regions we need to test.

What's the difference between < and ≤ in the final answer?

Since we want f(x) < 0 (strictly less than), we use strict inequality signs. The points x = -7 and x = 7 make f(x) = 0, so they're not included in our solution.

How can I visualize this problem?

Picture an upside-down parabola with vertex at (0, 49) crossing the x-axis at x = -7 and x = 7. The function is negative where the graph dips below the x-axis - that's outside the interval!

What if I get confused about which way the parabola opens?

Look at the coefficient of x²! If it's negative (like -1 here), the parabola opens downward. If it's positive, it opens upward. This determines where the function is positive vs negative.

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