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Which of these is an exponential function? \( y=x^{2}+\frac{1}{2} \quad y=\left(\frac{1}{4}\right)^{x} \)

Ask by Chadwick West. in the United States
Jan 24,2025

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The second function, \( y = \left(\frac{1}{4}\right)^x \), is an exponential function.

Solución

An **exponential function** is characterized by having the variable in the exponent. The general form of an exponential function is: \[ y = a \cdot b^x \] where: - \( a \) is a constant, - \( b \) is the base (with \( b > 0 \) and \( b \neq 1 \)), - \( x \) is the exponent. Let's analyze the given functions: 1. **\( y = x^2 + \frac{1}{2} \)** - Here, \( x \) is the base raised to the power of 2, and then a constant is added. - This is a **polynomial function**, not an exponential function, because the variable \( x \) is not in the exponent. 2. **\( y = \left(\frac{1}{4}\right)^x \)** - In this case, \( \frac{1}{4} \) is the base, and \( x \) is in the exponent. - This fits the form of an exponential function perfectly. **Conclusion:** The function \( y = \left(\frac{1}{4}\right)^x \) is an exponential function. **Answer:** Only the second function, \( y = \left(\dfrac{1}{4}\right)^{\,x} \), is an exponential function.

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Beyond the Answer

The function \( y=\left(\frac{1}{4}\right)^{x} \) is an exponential function because it has the form \( y = a^{x} \), where \( a \) is a constant. In this case, \( \frac{1}{4} \) is the base raised to the power of \( x \), which defines it as exponential. On the other hand, \( y=x^{2}+\frac{1}{2} \) is a polynomial function, not exponential. Exponential functions are not just about rapid growth; they can also model decay, like radioactivity! A classic example is the famous equation \( N(t) = N_0 e^{-\lambda t} \), which describes how a radioactive substance decreases over time. Now that’s some real-world drama, isn’t it? If you’re diving deeper into functions, consider spending time with resources on calculus and graphing techniques. Understanding concepts like the natural exponential function \( e \) and its properties can elevate your math game. A dazzling place to start could be classic textbooks or interactive online platforms that offer calculus problems and explanations.

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