Which Of The Following Is A Point On The Graph Of Y = ( 1 2 ) X Y=\left(\frac{1}{2}\right)^x Y = ( 2 1 ​ ) X ?A. (2, 1 4 \frac{1}{4} 4 1 ​ ) B. (0, 0) C. (0, 1 2 \frac{1}{2} 2 1 ​ ) D. (2, 1)

Which of the Following is a Point on the Graph of $y=\left(\frac{1}{2}\right)^x$?

Understanding the Graph of Exponential Functions

Exponential functions are a fundamental concept in mathematics, and their graphs play a crucial role in various mathematical and real-world applications. In this article, we will explore the graph of the function $y=\left(\frac{1}{2}\right)^x$ and determine which of the given points lies on this graph.

The General Form of Exponential Functions

The general form of an exponential function is $y=a^x$, where $a$ is a positive real number and $x$ is the variable. In this case, the function is $y=\left(\frac{1}{2}\right)^x$, where $a=\frac{1}{2}$. This function represents a decreasing exponential function, as the base $\frac{1}{2}$ is less than 1.

Properties of Exponential Functions

Exponential functions have several important properties that are essential to understand their behavior. Some of the key properties include:

• Domain and Range: The domain of an exponential function is all real numbers, and the range is all positive real numbers.
• Asymptotes: Exponential functions have a horizontal asymptote at $y=0$, as the function approaches 0 as $x$ approaches negative infinity.
• Increasing or Decreasing: If the base $a$ is greater than 1, the function is increasing. If the base $a$ is less than 1, the function is decreasing.

Graph of $y=\left(\frac{1}{2}\right)^x$

The graph of $y=\left(\frac{1}{2}\right)^x$ is a decreasing exponential function. As $x$ increases, the value of $y$ decreases. The graph has a horizontal asymptote at $y=0$ and passes through the point $(0, 1)$.

Evaluating the Given Points

Now, let's evaluate the given points to determine which one lies on the graph of $y=\left(\frac{1}{2}\right)^x$.

A. (2, $\frac{1}{4}$)

To determine if this point lies on the graph, we need to substitute $x=2$ and $y=\frac{1}{4}$ into the equation $y=\left(\frac{1}{2}\right)^x$.

$$\frac{1}{4}=\left(\frac{1}{2}\right)^2$$

This equation is true, as $\left(\frac{1}{2}\right)^2=\frac{1}{4}$. Therefore, the point (2, $\frac{1}{4}$) lies on the graph of $y=\left(\frac{1}{2}\right)^x$.

B. (0, 0)

To determine if this point lies on the graph, we need to substitute $x=0$ and $y=0$ into the equation $y=\left(\frac{1}{2}\right)^x$.

$$0=\left(\frac{1}{2}\right)^0$$

This equation is not true, as $\left(\frac{1}{2}\right)^0=1$. Therefore, the point (0, 0) does not lie on the graph of $y=\left(\frac{1}{2}\right)^x$.

C. (0, $\frac{1}{2}$)

To determine if this point lies on the graph, we need to substitute $x=0$ and $y=\frac{1}{2}$ into the equation $y=\left(\frac{1}{2}\right)^x$.

$$\frac{1}{2}=\left(\frac{1}{2}\right)^0$$

This equation is not true, as $\left(\frac{1}{2}\right)^0=1$. Therefore, the point (0, $\frac{1}{2}$) does not lie on the graph of $y=\left(\frac{1}{2}\right)^x$.

D. (2, 1)

To determine if this point lies on the graph, we need to substitute $x=2$ and $y=1$ into the equation $y=\left(\frac{1}{2}\right)^x$.

$$1=\left(\frac{1}{2}\right)^2$$

This equation is not true, as $\left(\frac{1}{2}\right)^2=\frac{1}{4}$. Therefore, the point (2, 1) does not lie on the graph of $y=\left(\frac{1}{2}\right)^x$.

Conclusion

In conclusion, the point (2, $\frac{1}{4}$) lies on the graph of $y=\left(\frac{1}{2}\right)^x$. The other points do not lie on the graph, as they do not satisfy the equation $y=\left(\frac{1}{2}\right)^x$.
Q&A: Understanding the Graph of $y=\left(\frac{1}{2}\right)^x$

Frequently Asked Questions

In this article, we will address some of the most frequently asked questions about the graph of $y=\left(\frac{1}{2}\right)^x$.

Q: What is the domain of the function $y=\left(\frac{1}{2}\right)^x$?

A: The domain of the function $y=\left(\frac{1}{2}\right)^x$ is all real numbers, as the base $\frac{1}{2}$ is a positive real number.

Q: What is the range of the function $y=\left(\frac{1}{2}\right)^x$?

A: The range of the function $y=\left(\frac{1}{2}\right)^x$ is all positive real numbers, as the function is always positive.

Q: What is the horizontal asymptote of the graph of $y=\left(\frac{1}{2}\right)^x$?

A: The horizontal asymptote of the graph of $y=\left(\frac{1}{2}\right)^x$ is $y=0$, as the function approaches 0 as $x$ approaches negative infinity.

Q: Is the graph of $y=\left(\frac{1}{2}\right)^x$ increasing or decreasing?

A: The graph of $y=\left(\frac{1}{2}\right)^x$ is decreasing, as the base $\frac{1}{2}$ is less than 1.

Q: What is the point of intersection of the graph of $y=\left(\frac{1}{2}\right)^x$ and the x-axis?

A: The point of intersection of the graph of $y=\left(\frac{1}{2}\right)^x$ and the x-axis is $(0, 1)$, as the function passes through this point.

Q: How can I determine if a point lies on the graph of $y=\left(\frac{1}{2}\right)^x$?

A: To determine if a point lies on the graph of $y=\left(\frac{1}{2}\right)^x$, substitute the x and y values into the equation $y=\left(\frac{1}{2}\right)^x$ and check if the equation is true.

Q: What is the significance of the graph of $y=\left(\frac{1}{2}\right)^x$ in real-world applications?

A: The graph of $y=\left(\frac{1}{2}\right)^x$ has various real-world applications, including modeling population growth, chemical reactions, and financial investments.

Q: How can I graph the function $y=\left(\frac{1}{2}\right)^x$ using a graphing calculator?

A: To graph the function $y=\left(\frac{1}{2}\right)^x$ using a graphing calculator, enter the function into the calculator and adjust the window settings to view the graph.

Conclusion

In conclusion, the graph of $y=\left(\frac{1}{2}\right)^x$ is a decreasing exponential function with a horizontal asymptote at $y=0$. The domain of the function is all real numbers, and the range is all positive real numbers. The graph passes through the point $(0, 1)$ and has various real-world applications. By understanding the properties and behavior of the graph of $y=\left(\frac{1}{2}\right)^x$, you can better analyze and solve problems involving exponential functions.