Solve f(x) = x² - 6x + 8 < 0: Graphical Analysis with Linear Function

Quadratic Inequalities with Interval Analysis

The following functions are graphed below:

f(x)=x26x+8 f(x)=x^2-6x+8

g(x)=4x17 g(x)=4x-17

For which values of x is
f(x)<0 f(x)<0 true?

BBBAAAKKK

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 For which values is the function negative?
00:04 We want to find the intersection points with the X-axis
00:11 We'll use the shortened multiplication formulas
00:17 Let's find the intersection points with the X-axis
00:22 Let's find the negative domains of the function, according to the graph
00:33 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

The following functions are graphed below:

f(x)=x26x+8 f(x)=x^2-6x+8

g(x)=4x17 g(x)=4x-17

For which values of x is
f(x)<0 f(x)<0 true?

BBBAAAKKK

2

Step-by-step solution

To solve the inequality f(x)=x26x+8<0 f(x) = x^2 - 6x + 8 < 0 , we follow these steps:

  • Step 1: Find the roots of the equation f(x)=0 f(x) = 0 using the factoring method.
  • Step 2: Determine the intervals formed by these roots.
  • Step 3: Test points from each interval to determine where f(x)<0 f(x) < 0 .

Step 1: Factor the quadratic equation x26x+8=0 x^2 - 6x + 8 = 0 .
Factoring gives: (x2)(x4)=0(x - 2)(x - 4) = 0.

Thus, the roots are x=2 x = 2 and x=4 x = 4 .

Step 2: The roots divide the number line into three intervals: x<2 x < 2 , 2<x<4 2 < x < 4 , and x>4 x > 4 .

Step 3: Choose a test point from each interval and plug it into f(x) f(x) :

  • For x<2 x < 2 , choose x=1 x = 1 : f(1)=126(1)+8=16+8=3 f(1) = 1^2 - 6(1) + 8 = 1 - 6 + 8 = 3 , which is positive, so f(x)0 f(x) \geq 0 .
  • For 2<x<4 2 < x < 4 , choose x=3 x = 3 : f(3)=326(3)+8=918+8=1 f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1 , which is negative, so f(x)<0 f(x) < 0 .
  • For x>4 x > 4 , choose x=5 x = 5 : f(5)=526(5)+8=2530+8=3 f(5) = 5^2 - 6(5) + 8 = 25 - 30 + 8 = 3 , which is positive, so f(x)0 f(x) \geq 0 .

Therefore, the interval where f(x)<0 f(x) < 0 is 2<x<4 2 < x < 4 .

The correct choice is:

2<x<4 2 < x < 4

3

Final Answer

2<x<4 2 < x < 4

Key Points to Remember

Essential concepts to master this topic
  • Factoring: Find roots by factoring x26x+8=(x2)(x4)=0 x^2 - 6x + 8 = (x-2)(x-4) = 0
  • Test Points: Choose x = 3 from interval (2,4): f(3)=1<0 f(3) = -1 < 0
  • Verification: Graph shows parabola below x-axis between roots x = 2 and x = 4 ✓

Common Mistakes

Avoid these frequent errors
  • Testing only one interval or misreading the inequality sign
    Don't just find the roots and guess the answer = wrong interval! Students often forget that parabolas change sign at each root, or mix up < and > symbols. Always test a point from each interval created by the roots to determine where the function is actually negative.

Practice Quiz

Test your knowledge with interactive questions

The following functions are graphed below:

\( f(x)=x^2-6x+8 \)

\( g(x)=4x-17 \)

For which values of x is
\( f(x)<0 \) true?

BBBAAAKKK

FAQ

Everything you need to know about this question

Why do I need to test points in each interval?

+

Because quadratic functions are parabolas that change from positive to negative (or vice versa) at each root. Testing points tells you exactly which intervals satisfy f(x)<0 f(x) < 0 .

How do I know which direction the parabola opens?

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Look at the coefficient of x2 x^2 ! Since it's positive (+1), the parabola opens upward, so it's negative between the roots and positive outside them.

What if I can't factor the quadratic easily?

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Use the quadratic formula: x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} . The roots will still divide the number line into intervals for testing.

Do I include the roots in my answer?

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No! Since we want f(x)<0 f(x) < 0 (strictly less than), the roots where f(x)=0 f(x) = 0 are not included. Use open intervals: 2<x<4 2 < x < 4 .

How does the graph help verify my answer?

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The graph shows where the parabola (blue curve) is below the x-axis. This visual confirmation matches our algebraic solution: between x = 2 and x = 4.

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