Question

oak wilt is a fungal disease that infects oak trees. scientists have discovered that forest is infected with oak wilt. they determined that they can use this exponent number of trees in the forest that will be infected after t years. f(t) = e^{0.4t} part a question graph the function f(t) = e^{0.4t}.

Explanation:

Step1: Identify the type of function

The function \( f(t) = e^{0.4t} \) is an exponential growth function since the exponent has a positive coefficient (\( 0.4>0 \)). The general form of an exponential function is \( y = ae^{kt} \), where \( a = 1 \) (since there's no coefficient in front of \( e^{0.4t} \)) and \( k=0.4 \).

Step2: Find the y - intercept

To find the y - intercept, we set \( t = 0 \). Substitute \( t = 0 \) into the function: \( f(0)=e^{0.4\times0}=e^{0}=1 \). So the graph passes through the point \( (0, 1) \).

Step3: Analyze the end - behavior

• As \( t

ightarrow+\infty \): Since the exponent \( 0.4t
ightarrow+\infty \) as \( t
ightarrow+\infty \), and the exponential function \( y = e^{x} \) increases without bound as \( x
ightarrow+\infty \), we have \( f(t)=e^{0.4t}
ightarrow+\infty \) as \( t
ightarrow+\infty \).

• As \( t

ightarrow-\infty \): The exponent \( 0.4t
ightarrow-\infty \) as \( t
ightarrow-\infty \), and \( e^{x}
ightarrow0 \) as \( x
ightarrow-\infty \), so \( f(t)=e^{0.4t}
ightarrow0 \) as \( t
ightarrow-\infty \).

Step4: Plot some additional points (optional but helpful)

Let's choose a few values of \( t \) to find corresponding \( f(t) \) values:

• When \( t = 1 \): \( f(1)=e^{0.4\times1}\approx e^{0.4}\approx1.4918 \)
• When \( t = 2 \): \( f(2)=e^{0.4\times2}=e^{0.8}\approx2.2255 \)
• When \( t=- 1 \): \( f(-1)=e^{0.4\times(-1)}=e^{-0.4}\approx0.6703 \)

To graph the function:

1. Start by plotting the y - intercept \( (0,1) \).
2. Use the end - behavior and the additional points to sketch a smooth curve that passes through \( (0,1) \), increases as \( t \) increases, and approaches \( y = 0 \) as \( t

ightarrow-\infty \). The curve should be concave up (since the second derivative of \( e^{kt} \) is \( k^{2}e^{kt}>0 \) for all \( t \)) and pass through the points we calculated like \( (1, e^{0.4})\approx(1,1.49) \), \( (2, e^{0.8})\approx(2,2.23) \) and \( (- 1,e^{-0.4})\approx(-1,0.67) \).

Answer:

The graph of \( f(t)=e^{0.4t} \) is an exponential growth curve passing through \( (0,1) \), increasing as \( t \) increases, approaching \( y = 0 \) as \( t
ightarrow-\infty \), and having a concave - up shape with additional points such as \( (1, e^{0.4})\approx(1,1.49) \), \( (2, e^{0.8})\approx(2,2.23) \) and \( (-1,e^{-0.4})\approx(-1,0.67) \) on it. (If we were to describe the graph in a more visual - based way for a graphing tool, we can use the key features: y - intercept at (0,1), exponential growth, concave up, passing through the calculated points)