To find the $x$-intercept of the function $f(x) = 8\left(\frac{1}{2}\right)^x - 2$, we need to set $f(x) = 0$ and solve for $x$.

1. Set up the equation: $8\left(\frac{1}{2}\right)^x - 2 = 0$

2. Move $-2$ to the other side: $8\left(\frac{1}{2}\right)^x = 2$

3. Divide both sides by 8: $\left(\frac{1}{2}\right)^x = \frac{1}{4}$

4. Rewrite $\frac{1}{4}$ as a power of $\frac{1}{2}$: $\frac{1}{4} = \left(\frac{1}{2}\right)^2$

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

5. Since the bases are the same, set the exponents equal to each other: $x = 2$

Thus, the $x$-intercept of $f(x)$ is $(2, 0)$.

The correct answer is:

• (2, 0)

Would you like a further breakdown of any steps, or do you have any questions?

Related Questions:

1. How would you find the $y$-intercept of this function?
2. What happens to the function as $x \to \infty$?
3. How would you determine the range of $f(x)$?
4. How would you find the asymptote of $f(x)$?
5. What transformations have been applied to the basic function $f(x) = \left(\frac{1}{2}\right)^x$?

Tip:

When dealing with exponents, always check if you can express both sides of the equation with the same base to simplify the process.