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Magazine Chapter 4: Problem 29
Exer. 29-34: Sketch the graph of \(f\). $$ f(x)=\left|x^{2}-6 x+5\right| $$
Short Answer
Expert verified
The graph has its roots at \(x = 1\) and \(x = 5\), with an inverted quadratic between them.
Step by step solution
01
Identify the Inner Function
Identify the quadratic function inside the absolute value, which is \( x^2 - 6x + 5 \). We will first analyze this without the absolute value.
02
Find the Roots of the Quadratic
Solve \( x^2 - 6x + 5 = 0 \) to find the roots using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Here, \( a=1, b=-6, \text{ and } c=5 \). This gives \( x = 1 \) and \( x = 5 \).
03
Analyze the Intervals
The roots divide the real line into intervals: \(( -\infty, 1), (1, 5), \text{ and } (5, \infty)\). Analyze the sign of \( x^2 - 6x + 5 \) in these intervals.
04
Determine Sign of Quadratic in Each Interval
- For \( x < 1 \), choose \( x = 0 \): \( 0^2 - 6(0) + 5 = 5 > 0 \). - For \( 1 < x < 5 \), choose \( x = 3 \): \( 3^2 - 6(3) + 5 = -4 \). - For \( x > 5 \), choose \( x = 6 \): \( 6^2 - 6(6) + 5 = 5 > 0 \).
05
Apply Absolute Value
Use the sign analysis to apply the absolute value:- For \( x < 1 \) and \( x > 5 \), \( |x^2 - 6x + 5| = x^2 - 6x + 5 \).- For \( 1 < x < 5 \), \( |x^2 - 6x + 5| = -(x^2 - 6x + 5) = -x^2 + 6x - 5 \).
06
Sketch the Graph
Based on the intervals:- For \(( -\infty, 1)\) and \((5, \infty)\), plot \( x^2 - 6x + 5 \) as a positive quadratic.- Between \( 1 \) and \( 5 \), plot the inverted quadratic \( -x^2 + 6x - 5 \).- The graph should meet smoothly at \( x = 1 \) and \( x = 5 \), where the function is zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Function
A **quadratic function** is a fundamental concept in mathematics, often represented by the general form \( ax^2 + bx + c \). Here, "\( a \)," "\( b \)," and "\( c \)" are constants, where \( a eq 0 \). The graph of a quadratic function is a curve called a **parabola**. Understanding parabolas is crucial as they have various real-life applications, such as designing satellite dishes or bridges. For the quadratic \( x^2 - 6x + 5 \), we can identify the components: \( a = 1 \), \( b = -6 \), and \( c = 5 \). This specific example leads to a parabola opening upwards because the leading coefficient "\( a \)" is positive.
- The roots are found where the quadratic equals zero, which is essential in determining where the graph crosses the x-axis.
- Roots provide vital points used to analyze the entire function, especially when combined with absolute values, as shown in \( f(x) = |x^2 - 6x + 5| \).
Graph Sketching
**Graph sketching** involves plotting the essential features of a function on a coordinate plane. When sketching the graph of an absolute value quadratic function, such as \( f(x) = |x^2 - 6x + 5| \), a clear understanding of its behavior is necessary. Start with identifying the basic quadratic function, \( x^2 - 6x + 5 \), and then recognize the effects of the absolute value. The problem becomes more interesting because absolute value affects how the graph parts below the x-axis are reflected upwards.
Sketch two parts:
- Initially, calculate and plot the roots, \( x = 1 \) and \( x = 5 \), as these are points where the parabola intersects the x-axis and determine intervals to consider.
- Plot the vertex of the parabola, which, because of the absolute value transform, will affect the portions of the curve under the x-axis to聽flip above it.
Sketch two parts:
- Between \(( -\infty, 1)\) and \((5, \infty)\), the parabola remains standard as \( x^2 - 6x + 5 \).
- Between \( 1 \) and \( 5 \), plot \( -x^2 + 6x - 5 \), the inverted portion.
Sign Analysis
Performing a **sign analysis** is pivotal in exploring how a function behaves across different intervals. For instance, with \( x^2 - 6x + 5 \), recognize which intervals make it positive or negative to handle the absolute value more effectively. This analysis decides the output sign of the absolute value part.
- Divide the real number line by the roots identified from the quadratic, at \( x = 1 \) and \( x = 5 \), resulting in intervals \(( -\infty, 1)\), \((1, 5)\), and \((5, \infty)\).
- Check points within each interval to confirm if the quadratic part is positive or negative.
- In the first interval \(( -\infty, 1)\), as all calculations show positivity, the integrity of the quadratic part is retained.
- In the second \((1, 5)\), negativity necessitates flipping the quadratic part due to absolute value, switching it to opposite signs.
- Finally, in \((5, \infty)\), it repeats positivity, so it remains unchanged.
