Question

Use transformations to graph each function. Determine the domain, range, horizontal asymptote, and y-intercept of each function. $$ f(x)=2^{x+2} $$

Short answer

Domain: \( (-\forall, + \forall) \). Range: \( (0, + \forall) \). Horizontal Asymptote: \( y = 0 \). Y-Intercept: \( (0, 4) \).

Step-by-step solution

Identify the Base Function

The base function is the exponential function \(2^x\). It has a horizontal asymptote at \(y=0\), a y-intercept at \( (0, 1) \), a domain of all real numbers \( (-\forall, + \forall) \), and a range of \( (0, + \forall) \).

Determine the Transformation

The given function is \(f(x) = 2^{x+2}\). This represents a horizontal shift 2 units to the left, because adding 2 inside the function moves the graph left by 2 units.

Apply the Horizontal Shift

Shift the graph of \(2^x\) 2 units to the left to get the graph of \(2^{x+2}\). The horizontal shift does not affect the horizontal asymptote, y-intercept, domain, or range.

Determine the Domain

The domain of \(f(x) = 2^{x+2}\) remains all real numbers \((- \forall, + \forall)\) because an exponential function is defined for all x-values.

Determine the Range

The range of \(f(x) = 2^{x+2}\) remains \( (0, + \forall) \) since exponential functions only produce positive values.

Find the Horizontal Asymptote

The horizontal asymptote for \(f(x) = 2^{x+2}\) remains \(y=0\), as shifting left or right does not change the asymptote of an exponential function.

Find the Y-Intercept

To find the y-intercept, plug in \(x=0\) into \(f(x) = 2^{x+2}\). This gives \(f(0) = 2^{0+2} = 2^2 = 4\). Thus, the y-intercept is at \((0, 4)\).

Key concepts

Transformations

Graphing exponential functions often involves transformations. The base function here is the exponential function \(2^x\). To graph the given function \(f(x) = 2^{x+2}\), we need to understand how transformations affect it. Adding 2 inside the exponent represents a horizontal shift to the left by 2 units.
Here's why:

• If we add a constant \(h\) inside the function, \(f(x) = 2^{x+h}\), it shifts the graph horizontally. Positive values of \(h\) shift it to the left, while negative values shift it to the right.
• In our problem, \((x+2)\) means we shift left by 2 units.

This is why the graph of \(f(x) = 2^{x+2}\) looks just like \(2^x\) but moved 2 units left.
The x-intercepts and y-intercepts move accordingly, but many properties remain unchanged.

Domain and Range

The domain and range of exponential functions are very predictable. Understanding these can make graphing easier.
The domain of an exponential function like \(2^x\) is all real numbers, \(-\forall < x < +\forall\).

• This means you can plug in any x-value and get a corresponding y-value.
• Since transformations like shifts do not affect the domain, \((x+2)\) still allows any x-value.

The range of \(2^x\) is \((0, + \forall)\):

• Exponential functions only produce positive values because a positive base raised to any power never equals zero or a negative number.
• Thus, even after shifting horizontally, the range remains unchanged.

Understanding these ensures we know that the graph is defined everywhere and only outputs positive numbers.

Horizontal Asymptote

Exponential functions have a horizontal asymptote. This is a line that the graph approaches but never touches.
For the base function \(2^x\), the horizontal asymptote is at y=0. No matter how large or small x gets, \(2^x\) never actually reaches zero.

• Transformations like horizontal shifts do not change this property.
• Therefore, for \((x+2)\) in \(2^{x+2}\), the horizontal asymptote still remains at y=0.

This is a consistent feature of exponential growth or decay, signifying that as x goes to negative infinity, the function value approaches zero.

Y-Intercept

The y-intercept of a function is the point where the graph crosses the y-axis. For exponential functions, you find this by setting x to 0.

• For the base function \(2^x\), plugging in x=0 gives \(2^0 = 1\). Hence, the y-intercept of \(2^x\) is (0,1).
• For our transformed function \(2^{x+2}\), setting x to 0 gives us \(2^{0+2} = 2^2 = 4\).

Thus, \(f(0) = 4\), meaning the y-intercept of \(f(x) = 2^{x+2}\) is (0, 4). This shift from (0, 1) to (0, 4) is directly due to the horizontal shift applied to the exponent in the function.
The y-intercept, therefore, moves predictably with transformations inside the exponent.