Question

Use the table to answer the question. x-Values y-Values 0 5 1 15 2 45 3 135 4 405 Select the function that best models the data and determine the value of the function at x=−1 . (1 point) Responses f(x)=2.23⋅(1.40)x ; f(−1)=1.59 f of x is equal to 2 point 2 3 times 1 point 4 0 to the x th power ; f of negative 1 is equal to 1 point 5 9 f(x)=5⋅3x ; f(−1)=53 f of x is equal to 5 times 3 to the x th power; f of negative 1 is equal to 5 thirds f(x)=92x−63 ; f(−1)=−155 f of x is equal to 92 x minus 63; f of negative 1 is equal to negative 155 f(x)=41.43x2−73.71x+19.86 ; f(−1)=135

Answer 1

To determine which function best models the given data, we can look for patterns in the x and y values:

• When \( x = 0 \), \( y = 5 \)
• When \( x = 1 \), \( y = 15 \)
• When \( x = 2 \), \( y = 45 \)
• When \( x = 3 \), \( y = 135 \)
• When \( x = 4 \), \( y = 405 \)

Let's analyze these values:

1. Look for a pattern:
   • The values \( y \) seem to be multiplying by 3 as \( x \) increases by 1.
   • \( 5 \times 3 = 15 \)
   • \( 15 \times 3 = 45 \)
   • \( 45 \times 3 = 135 \)
   • \( 135 \times 3 = 405 \)

From this, we can determine that the function is likely exponential in the form of \( f(x) = a \cdot 3^x \).

2. Find the value of \( a \):

   • To find \( a \), we can use the y-value at \( x = 0 \): \[ f(0) = a \cdot 3^0 = a = 5 \] Therefore, the function is: \[ f(x) = 5 \cdot 3^x \]

3. Calculate \( f(-1) \): \[ f(-1) = 5 \cdot 3^{-1} = 5 \cdot \frac{1}{3} = \frac{5}{3} \]

Now, let's compare this with the provided options:

• The correct function that fits the data is \( f(x) = 5 \cdot 3^x \) and \( f(-1) = \frac{5}{3} \).

Thus, the response is: f(x)=5⋅3^x ; f(−1)=\frac{5}{3}