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Chapter 4: Problem 28

Sketch the graph of each function. $$h(x)=-5(3)^{x}$$

Short Answer

Expert verified
The graph of the function \(h(x)=-5 \cdot 3^x\) starts at the y-intercept point (0,-5). Since the function is a negative exponential, the graph decreases rapidly from (0, -5) and then approaches the x-axis as 'x' increases.

Step by step solution

01

Identify Basic Form

Recognize that the equation \(h(x)=-5 \cdot 3^x\) is in the form of \(y=ab^x\). Here, 'a' is -5 and 'b' is 3.
02

Determine Y-Intercept

The y-intercept is found when 'x' equals 0. Plugging 'x' as 0 into the function equation, \(h(0) = -5 \cdot 3^0 = -5\), we find that the y-intercept is -5.
03

Identify Horizontal Asymptote

The graph never touches the horizontal line \(y = 0\), but it gets arbitrarily close as 'x' tends to infinity. This line is called a horizontal asymptote.
04

Determine the Behavior and Sketch the Graph

Since the 'a' value is negative, the usual growth pattern of an exponential function is reflected across the x-axis. We have a decreasing type exponential function. Start at the intercept and draw a curve that approaches the line \( y = 0\) as \( x \rightarrow \infty \) and the line \( y = -5 \) becomes an upper bound as \( x \rightarrow -\infty\).

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Key Concepts

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

Graph Sketching
Graph sketching involves drawing a rough graph of a function by understanding its key features. To sketch the graph of an exponential function like \( h(x) = -5 \, (3)^x \), we start by identifying important points and behaviors.
  • Identify the Equation Form: Recognize the function is in the form \( y = ab^x \). Here, \( a = -5 \) and \( b = 3 \).
  • Determine the Y-Intercept: Plugging \( x = 0 \) into the equation gives \( h(0) = -5 \times 3^0 = -5 \). So, the graph crosses the y-axis at -5.
Graph sketching is a crucial skill because it helps visualize how values of \( x \) affect the function. When sketching, note that this graph reflects across the x-axis because of the negative sign in front of 5. Always label key points like the y-intercept and mark the direction they approach horizontal asymptotes.
Horizontal Asymptote
Horizontal asymptotes offer insight into the behavior of a graph at extreme values of \( x \). With the function \( h(x) = -5 (3)^x \), this graph has a horizontal asymptote at \( y = 0 \).
  • Understanding Asymptotic Behavior: As \( x \) becomes very large, \( 3^x \) increases exponentially, but the negative sign before 5 causes \( h(x) \) to decrease and get very close to zero.
  • Graph Never Touches Asymptote: Despite approaching \( y = 0 \), the graph will never actually touch this line.
This concept is key in exponential functions, as it helps predict the end behavior of a graph. Identifying horizontal asymptotes is essential for defining the sketch boundaries visually.
Exponential Decay
Exponential decay describes a process where quantities reduce at a consistent percentage rate over time. The function \( h(x) = -5 \times 3^x \) demonstrates exponential decay when reflected due to the negative coefficient.
  • Reflecting Across the X-axis: Typically, an exponential function like \( b = 3 \) would imply growth. However, the negative sign flips the growth to decay.
  • Visualizing Decay: As \( x \) increases, \( h(x) \) decreases towards the horizontal asymptote at \( y = 0 \).
Exponential decay is widely observed in real-life scenarios such as radioactive decay or depreciation of value. It's important to distinguish when a function is increasing versus decreasing to correctly interpret associated applications.

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Most popular questions from this chapter

Plutonium is a radioactive element that has a half-life of 24,360 years. The half-life of a radioactive substance is the time it takes for half of the substance to decay (which means the other half will still exist after that length of time). Find an exponential function of the form \(f(t)=A e^{k t}\) that gives the amount of plutonium left after \(t\) years if the initial amount of plutonium is 10 pounds. How long will it take for the plutonium to decay to 2 pounds?

Evaluate the given quantity by referring to the function \(f\) given in the following table. $$\begin{array}{cc}x & f(x) \\\\-2 & 1 \\\\-1 & 2 \\\0 & 0 \\\1 & -1 \\\2 & -2\end{array}$$ $$f^{-1}(1)$$

Find the inverse of the given function. Then graph the given function and its inverse on the same set of axes. $$g(x)=\frac{-1}{2 x}$$

The spread of a disease can be modcled by a logistic function. For example, in carly 2003 there was an outbreak of an illness called SARS (Severe Acute Respiratory Syndrome) in many parts of the world. The following table gives the total momber of cases in Canada for the wecks following March 20,2003 (Source: World Health Organization) (Note: The total number of cases dropped from 149 to 140 between weeks 3 and 4 because some of the cases thought to be SARS were reclassified as other discases.) $$\begin{array}{|c|c|}\hline\text { Weeks since } & \\\\\text { March } 20,2003 & \text { Total Cases } \\\0 & 9 \\\1 & 62 \\\2 & 132 \\\3 & 149 \\\4 & 140 \\\5 & 216 \\\6 & 245 \\\7 & 252 \\\8 & 250 \\\\\hline\end{array}$$ (a) Explain why a logistic function would suit this data well. (b) Make a scatter plot of the data and find the logistic function of the form \(f(x)=\frac{\epsilon}{1+a \varepsilon^{-1}}\) that best fits the data. (c) What docs \(c\) signify in your model? (d) The World Health Organization declared in July 2003 that SARS no longer posed a threat in Canada. By analyring this data, explain why that would be so.

Two students have an argument. One says that the inverse of the function \(f\) given by the expression \(f(x)=6\) is the function \(g\) given by the expression \(g(x)=\frac{1}{6} ;\) the other claims that \(f\) has no inverse. Who is correct and why?
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