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

Q: Why do I factor the quadratic first?

Factoring reveals the roots (where the parabola crosses the x-axis). These roots divide the number line into intervals where the function keeps the same sign - either all positive or all negative.

Q: Why isn't the answer 2 ≤ x ≤ 4?

The inequality is strictly less than zero (\( f(x) ), not less than or equal to. At x = 2 and x = 4, the function equals exactly zero, not negative.

Q: What if I can't factor the quadratic?

Use the quadratic formula to find the roots first: $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $. Then create your intervals and test points the same way.

Q: How can I visualize this on a graph?

Look for where the parabola dips below the x-axis (negative y-values). This happens between the two x-intercepts for upward-opening parabolas like this one.

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:10 Let's find the negative domain of this function!

00:14 The negative domain is where the graph is below the X axis.

00:18 Now, we'll find where the graph meets the X axis. These are the intersection points.

00:24 Next, let's break it down into a trinomial expression.

00:28 Here is the trinomial we need. Looks good!

00:32 Now, let's find the solutions for this trinomial.

00:35 Great! These are where the graph crosses the X axis.

00:40 Let's determine where the function dips below the X axis.

00:44 And that's how we find the solution to the question. Well done!

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

The following function is graphed below:

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

For which values of x is

$f(x)<0$ true?

2

Step-by-step solution

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 $2 < x < 4$ .

3

Final Answer

$2 < x < 4$

Key Points to Remember

Essential concepts to master this topic

Factoring: Set quadratic equal to zero to find roots
Technique: Test signs in intervals: (x-2)(x-4) = (+)(−) = (−) when 2 < x < 4
Check: Pick x=3: (3-2)(3-4) = (1)(-1) = -1 < 0 ✓

Common Mistakes

Avoid these frequent errors

Testing the wrong inequality direction
Don't solve where f(x) > 0 when asked for f(x) < 0 = opposite answer! This gives you the intervals where the parabola is above the x-axis instead of below. Always check whether you need the function to be positive or negative.

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

Why do I factor the quadratic first?

+

Factoring reveals the roots (where the parabola crosses the x-axis). These roots divide the number line into intervals where the function keeps the same sign - either all positive or all negative.

How do I know which interval to choose?

+

Test a point from each interval! Pick any x-value between the roots and substitute it into the factored form. If the result is negative, that interval works for $f(x) < 0$ .

Why isn't the answer 2 ≤ x ≤ 4?

+

The inequality is strictly less than zero ( $f(x) < 0$ ), not less than or equal to. At x = 2 and x = 4, the function equals exactly zero, not negative.

What if I can't factor the quadratic?

+

Use the quadratic formula to find the roots first: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . Then create your intervals and test points the same way.

How can I visualize this on a graph?

+

Look for where the parabola dips below the x-axis (negative y-values). This happens between the two x-intercepts for upward-opening parabolas like this one.

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

Quadratic Inequalities with Factored Form

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