Determining Negative Values for the Quadratic Function: Where Is y = x² - 6x + 8 Less Than Zero?

The following function is graphed below:

$y=x^2-6x+8$

For which values of x is

$f(x)<0$ true?

To determine where $f(x) = x^2 - 6x + 8$ is less than zero, we need to find the roots of the quadratic equation and test the intervals determined by them.

Step 1: Factor the quadratic.
The equation can be rewritten as $x^2 - 6x + 8 = (x - 2)(x - 4) = 0$.
Thus, the roots are $x = 2$ and $x = 4$.

Step 2: Using these roots, we can identify intervals to test where the product $(x - 2)(x - 4) < 0$. The intervals derived from the roots are:

• Interval 1: $x < 2$
• Interval 2: $2 < x < 4$
• Interval 3: $x > 4$

Step 3: Test each interval to find where $f(x) < 0$.
- For $x < 2$, both factors $(x - 2)$ and $(x - 4)$ are negative, thus their product is positive.
- For $2 < x < 4$, $(x - 2)$ is positive, and $(x - 4)$ is negative, so their product is negative.
- For $x > 4$, both $(x - 2)$ and $(x - 4)$ are positive, so their product is positive.

Therefore, the function $f(x) = x^2 - 6x + 8$ is less than zero for the range $2 < x < 4$.

Thus, the values for which $f(x) < 0$ is true are $2 < x < 4$.