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Chapter 3: Problem 83

(a) Graph \(f(x)=|x-3|-3\) using transformations. (b) Find the area of the region that is bounded by \(f\) and the \(x\) -axis and lies below the \(x\) -axis.

Short Answer

Expert verified
The area of the region below the x-axis is 9 square units.

Step by step solution

01

Understand the Base Graph

The base graph here is the absolute value function, denoted as \( g(x) = |x| \). This graph forms a 'V' shape with its vertex at the origin (0,0).
02

Apply Horizontal Shift

The function \( f(x) = |x-3| - 3 \) includes a horizontal shift. The term \( x-3 \) indicates a shift 3 units to the right. Therefore, the vertex of the 'V' shape now moves to (3,0).
03

Apply Vertical Shift

Next, note the \( -3 \) term outside the absolute value. This causes a vertical shift downward by 3 units. The new vertex of the graph becomes (3, -3).
04

Plot the Transformed Graph

Using the transformations, plot the function \( f(x) = |x-3| - 3 \) by starting at the new vertex (3, -3) and drawing the usual 'V' shape.
05

Determine Where the Graph Crosses the x-axis

To find intersections with the x-axis, set \( f(x) = 0 \):\[ |x-3| - 3 = 0 \Rightarrow |x-3| = 3 \Rightarrow x - 3 = 3 \text{ or } x - 3 = -3 \Rightarrow x = 6 \text{ or } x = 0 \]. Thus, the graph crosses the x-axis at x = 6 and x = 0.
06

Calculate the Bounded Area Below the x-axis

The region below the x-axis and bounded by the graph can be visualized as an inverted triangle with base from x = 0 to x = 6. The height of the triangle is the distance from the x-axis to the lowest point of the graph, which is 3 units down (from y = 0 to y = -3). Calculate the area: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3 = 9 \text{ square units} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

absolute value function
The absolute value function, denoted as \( g(x) = |x| \), is fundamental in understanding graph transformations. This function creates a 'V' shaped graph, symmetric about the y-axis. The vertex, or the tip of the 'V', is positioned at the origin (0,0). The graph reflects across the y-axis, making it identical on both sides. When dealing with absolute value functions, keep in mind its unique property: it always outputs non-negative values for any input. This property is key in understanding how transformations affect the graph.
horizontal shift
A horizontal shift changes the position of the graph left or right along the x-axis. It's represented inside the function argument. For instance, in the function \( f(x) = |x-3| - 3 \), the term \( x-3 \) means the graph of \( |x| \) shifts 3 units to the right. Think of it as relocating the starting point of your graph. This transformation does not alter the shape of the graph; it simply moves the vertex. By comparing \( |x| \) and \( |x-3| \), you'll see the 'V' shape has slid from (0,0) to (3,0).
vertical shift
A vertical shift moves the graph up or down along the y-axis. It's reflected as a constant term added or subtracted outside the function. For the function \( f(x) = |x-3| - 3 \), the \( -3 \) outside the absolute value function indicates a vertical shift downward by 3 units. This transformation alters the y-value of the vertex, moving it from (3,0) to (3, -3). Remember, the shape remains unchanged; we just adjust the graph鈥檚 position vertically. Imagine lifting or lowering the entire 'V' shape along the y-axis: it now starts from (3,-3).
area under curve
Finding the area under a curve involves calculating the space between the graph and the x-axis within a range of x-values. For our function \( f(x) = |x-3| - 3 \), the area of interest is below the x-axis, forming a triangle between x = 0 and x = 6. The graph crosses the x-axis at these points. The height of this inverted triangle is from y = 0 to y = -3, which is 3 units. We use the formula for the area of a triangle: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]. Here, base is the distance between x = 0 and x = 6, totaling 6 units. Thus, the area is \( \frac{1}{2} \times 6 \times 3 = 9 \) square units.

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Most popular questions from this chapter

The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. Problems \(85-92\) require the following discussion of a secant line. The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. \(f(x)=x^{2}+2 x\)

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