Solve y=-x²+10x-16: Finding Where Function is Negative

Look at the following function:

$y=-x^2+10x-16$

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

$f(x) < 0$

To determine where the function $f(x) = -x^2 + 10x - 16$ is less than zero, we should first find the roots by solving $f(x) = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = 10$, and $c = -16$, we can find the roots:

• Calculate the discriminant: $b^2 - 4ac = 10^2 - 4(-1)(-16) = 100 - 64 = 36$.
• Find the roots: $x = \frac{-10 \pm \sqrt{36}}{2(-1)} = \frac{-10 \pm 6}{-2}$.
• The roots are $x = \frac{-10 + 6}{-2} = 2$ and $x = \frac{-10 - 6}{-2} = 8$.

These roots divide the number line into intervals: $x < 2$, $2 < x < 8$, and $x > 8$.

To determine where $f(x) < 0$, test a point in each interval:

• For $x < 2$, choose $x = 0$. $f(0) = -0^2 + 10(0) - 16 = -16$ (which is less than zero).
• For $2 < x < 8$, choose $x = 5$. $f(5) = -(5)^2 + 10(5) - 16 = -25 + 50 - 16 = 9$ (which is greater than zero).
• For $x > 8$, choose $x = 10$. $f(10) = -(10)^2 + 10(10) - 16 = -100 + 100 - 16 = -16$ (which is less than zero).

Therefore, the function $f(x)$ is negative for $x < 2$ and $x > 8$.

Thus, the values of $x$ for which $f(x) < 0$ are $x < 2$ or $x > 8$.

The correct choice corresponding to this solution is: $x > 8$ or $x < 2$.