Which exponential function has an initial value of 3? x f(x) -2 -3.938 -1 -3.75 0 -3 1 0 2 12
y 8 6 4 2 x -3 -2 -1 1 2 3
Step 1: Check the table.
Looking at the table:
- f(0)=−3f(0) = -3f(0)=−3
This function does not have an initial value of 3, so the table is not correct.
Step 2: Check the equation f(x)=13(9)xf(x) = \frac{1}{3}(9)^xf(x)=31(9)x.
Let’s plug in x=0x = 0x=0:f(0)=13(9)0=13(1)=13f(0) = \frac{1}{3}(9)^0 = \frac{1}{3}(1) = \frac{1}{3}f(0)=31(9)0=31(1)=31
So, this function has an initial value of 13\frac{1}{3}31, not 3.
Step 3: Check the graph.
Look at where the graph crosses the y-axis. The y-axis is where x=0x = 0x=0.
From the graph, we see that when x=0x = 0x=0, y=3y = 3y=3.
So:f(0)=3f(0) = 3f(0)=3
That matches our condition.
Step 4: Check the equation f(x)=3xf(x) = 3^xf(x)=3x.
Plug in x=0x = 0x=0:f(0)=30=1f(0) = 3^0 = 1f(0)=30=1
So, this function also does not have an initial value of 3.
Final Conclusion:
Only the graph represents an exponential function with an initial value of 3. Therefore, it is the correct answer.
