Draw the graph of the function and state the domain and range. $$ f(x)=3^{x+2} $$

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Alright, I need to draw the graph of the function $$ f(x) = 3^{x+2} $$ and state its domain and range. Let's start by understanding the function itself.

First, $$ f(x) = 3^{x+2} $$ is an exponential function. The base here is 3, which is greater than 1, so the function will grow rapidly as $$ x $$ increases. The exponent is $$ x + 2 $$, which means the graph will be shifted 2 units to the left compared to the basic exponential function $$ 3^x $$.

Let me recall the general properties of exponential functions. For $$ f(x) = a^x $$, where $$ a > 1 $$, the function is always increasing, and its domain is all real numbers, while the range is all positive real numbers. Since our function is $$ 3^{x+2} $$, the domain should still be all real numbers because the exponent $$ x + 2 $$ can take any real value. The range should also be all positive real numbers, but I need to confirm if the shift affects this.

Wait, does the horizontal shift affect the range? No, because the exponential function's range is always positive regardless of horizontal shifts. So, the range remains $$ (0, \infty) $$.

Now, let's think about key points to plot. For $$ x = -2 $$, $$ f(-2) = 3^{-2 + 2} = 3^0 = 1 $$. So, the graph passes through the point $$ (-2, 1) $$. For $$ x = 0 $$, $$ f(0) = 3^{0 + 2} = 3^2 = 9 $$, so it passes through $$ (0, 9) $$. For $$ x = 2 $$, $$ f(2) = 3^{2 + 2} = 3^4 = 81 $$, so $$ (2, 81) $$ is another point.

Let me consider the behavior as $$ x $$ approaches negative infinity. As $$ x $$ becomes more negative, $$ x + 2 $$ also becomes more negative, and $$ 3^{x+2} $$ approaches 0. So, the graph approaches the x-axis (y = 0) as $$ x $$ approaches negative infinity.

As $$ x $$ approaches positive infinity, $$ x + 2 $$ becomes very large, and $$ 3^{x+2} $$ grows without bound. So, the graph rises steeply as $$ x $$ increases.

Now, to sketch the graph, I'll plot these points and draw a smooth curve that passes through them, approaching the x-axis on the left and rising rapidly on the right.

But wait, is there any symmetry or special features? Exponential functions are one-to-one, meaning they are always increasing and never decrease. So, the graph should be strictly increasing without any turning points.

Also, the y-intercept is at $$ (0, 9) $$, and the graph crosses the x-axis at $$ (-2, 1) $$. There are no horizontal asymptotes other than the x-axis, which the graph approaches but never touches.

Let me double-check the calculations:
- $$ f(-2) = 3^{0} = 1 $$ ✅
- $$ f(0) = 3^{2} = 9 $$ ✅
- $$ f(2) = 3^{4} = 81 $$ ✅

Everything seems correct. So, the graph of $$ f(x) = 3^{x+2} $$ is an exponential curve shifted 2 units to the left, passing through the points $$ (-2, 1) $$, $$ (0, 9) $$, and $$ (2, 81) $$, approaching the x-axis as $$ x $$ decreases and rising rapidly as $$ x $$ increases.

**Graph of $$ f(x) = 3^{x+2} $$:** - **Domain:** All real numbers ($$ (-\infty, \infty) $$) - **Range:** Positive real numbers ($$ (0, \infty) $$) **Key Points:** - $$ (-2, 1) $$ - $$ (0, 9) $$ - $$ (2, 81) $$ **Behavior:** - Approaches the x-axis ($$ y = 0 $$) as $$ x $$ approaches negative infinity. - Rises rapidly as $$ x $$ approaches positive infinity. - The graph is strictly increasing without any turning points.

Draw the graph of the function and state the domain and range. \( f(x)=3^{x+2} \)

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