1. **State the problem:** We are given the exponential function $f(x) = 8\left(\frac{3}{4}\right)^x$ and need to find its domain, range, asymptote, and end behavior. 2. **Recall the general form and properties:** The function is of the form $f(x) = a b^x$ where $a=8$ and $b=\frac{3}{4}$. Since $0 < b < 1$, this is an exponential decay function. 3. **Domain:** The domain of any exponential function $a b^x$ is all real numbers because $x$ can be any real number. $$\text{Domain} = (-\infty, \infty)$$ 4. **Range:** Since $a=8 > 0$ and $b^x > 0$ for all $x$, the function values are always positive but never zero. The function approaches zero but never reaches it. $$\text{Range} = (0, 8]$$ Note: At $x=0$, $f(0) = 8 \times 1 = 8$, so the maximum value is 8. 5. **Asymptote:** The horizontal asymptote is the value that $f(x)$ approaches as $x \to \infty$ or $x \to -\infty$. Since $b=\frac{3}{4} < 1$, as $x \to \infty$, $\left(\frac{3}{4}\right)^x \to 0$, so $$y = 0$$ is the horizontal asymptote. 6. **End behavior:** - As $x \to \infty$, $f(x) = 8\left(\frac{3}{4}\right)^x \to 8 \times 0 = 0$. - As $x \to -\infty$, $f(x) = 8\left(\frac{3}{4}\right)^x = 8 \times \left(\frac{4}{3}\right)^{-x} \to \infty$ because $\left(\frac{3}{4}\right)^x$ grows without bound when $x$ is very negative. **Summary:** - Domain: $(-\infty, \infty)$ - Range: $(0, 8]$ - Horizontal asymptote: $y=0$ - End behavior: $f(x) \to 0$ as $x \to \infty$, and $f(x) \to \infty$ as $x \to -\infty$.