Question

Graph the function y=5(14)x . Complete the ordered pairs for the x-values of −2 , −1 , 0, and 1 by finding the y-values.(1 point)
Responses

(−2,−2.5), (1,1.25), (0,0) and (1,1.2)
left parenthesis negative 2 comma negative 2.5 right parenthesis , left parenthesis 1 comma 1.25 right parenthesis , left parenthesis 0 comma 0 right parenthesis and left parenthesis 1 comma 1.2 right parenthesis

(−2,−2.99), (−1,−1.49), (0,0), and (1,1.49)
left parenthesis negative 2 comma negative 2.99 right parenthesis , left parenthesis negative 1 comma negative 1.49 right parenthesis , left parenthesis 0 comma 0 right parenthesis , and left parenthesis 1 comma 1.49 right parenthesis

(−2,0.45), (−1,0.67), (0,1), and (1,1.50)
left parenthesis negative 2 comma 0.45 right parenthesis , left parenthesis negative 1 comma 0.67 right parenthesis , left parenthesis 0 comma 1 right parenthesis , and left parenthesis 1 comma 1.50 right parenthesis

(−2,80), (−1,20), (0,5), and (1,1.25)

Answer 1

To find the y-values for the function \( y = 5(14)^x \) at the given x-values of -2, -1, 0, and 1, let's calculate each one:

1. For \( x = -2 \):
\[
y = 5(14)^{-2} = 5 \cdot \frac{1}{14^2} = 5 \cdot \frac{1}{196} = \frac{5}{196} \approx 0.0256
\]

2. For \( x = -1 \):
\[
y = 5(14)^{-1} = 5 \cdot \frac{1}{14} \approx 5 \cdot 0.0714 \approx 0.3571
\]

3. For \( x = 0 \):
\[
y = 5(14)^0 = 5 \cdot 1 = 5
\]

4. For \( x = 1 \):
\[
y = 5(14)^1 = 5 \cdot 14 = 70
\]

Now we have the ordered pairs:
- For \( x = -2 \), \( y \approx 0.0256 \)
- For \( x = -1 \), \( y \approx 0.3571 \)
- For \( x = 0 \), \( y = 5 \)
- For \( x = 1 \), \( y = 70 \)

Thus, the ordered pairs are approximately:
- \( (-2, 0.0256) \)
- \( (-1, 0.3571) \)
- \( (0, 5) \)
- \( (1, 70) \)

None of the provided options match the calculated values since they don't accurately represent the function's output at the specified input. If you need a specific choice from the answers provided, please check if there's either a mistake in the choices or in the evaluation of the function given.