/4 Points] DETAILS OSCAT1 6.2.073. MY NOTES The graph of $f(x)=16^{x}$ is reflected about the $x$-axis and shifted upward 3 units. What is the equation of the new function, $g(x)$ \[ g(x)= \] State the $y$-intercept of $g(x)$. \[ (x, y)=(\square) \] State the domain and range of $g(x)$. (Enter your answers using interval notation.) domain range
Solution
Step 1 ：The graph of the function $f(x)=16^{x}$ is reflected about the $x$-axis and shifted upward 3 units. This means we change the sign of the function and add 3 to it. Therefore, the new function $g(x)$ is $-16^{x} + 3$.
Step 2 ：The y-intercept of a function is the value of the function when $x = 0$. So, to find the y-intercept of $g(x)$, we substitute $x = 0$ into the equation of $g(x)$, which gives us $(0, 2)$.
Step 3 ：The domain of a function is the set of all possible x-values, and the range is the set of all possible y-values. Since $16^{x}$ is defined for all real numbers, the domain of $g(x)$ is all real numbers. The range of $g(x)$ is all real numbers less than or equal to 3, because the function $-16^{x}$ decreases as $x$ increases, and the function is shifted upward by 3 units.
Step 4 ：Final Answer: The equation of the new function, $g(x)$, is \(\boxed{g(x) = -16^{x} + 3}\). The $y$-intercept of $g(x)$ is \(\boxed{(0, 2)}\). The domain of $g(x)$ is \(\boxed{(-\infty, \infty)}\) and the range of $g(x)$ is \(\boxed{(-\infty, 3]}\).
