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

Look at the following function:

$y=-x^2+49$

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

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

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 $x < -7$