To determine which exponential equation contains the points \((-1, 5)\) and \((2, 5000)\), we can use the general form of an exponential function:
\[ y = a \cdot b^x \]
where \(a\) is a constant and \(b\) is the base of the exponential function.
Let's analyze each candidate equation:
-
For \(y = 500^x\):
- When \(x = -1\): \(y = 500^{-1} = \frac{1}{500} \neq 5\)
- When \(x = 2\): \(y = 500^2 = 250000 \neq 5000\)
- Not a match.
-
For \(y = 50 \cdot (-10)^x\):
- This equation will not yield real values for positive \(x\) since it involves negative bases raised to real numbers.
- Not a match.
-
For \(y = 50 \cdot 10^x\):
- When \(x = -1\): \(y = 50 \cdot 10^{-1} = 50 \cdot 0.1 = 5\) (Match for point (-1, 5))
- When \(x = 2\): \(y = 50 \cdot 10^2 = 50 \cdot 100 = 5000\) (Match for point (2, 5000))
- This is a match.
-
For \(y = -50 \cdot (10)^x\):
- When \(x = -1\): \(y = -50 \cdot (10)^{-1} = -50 \cdot 0.1 = -5 \neq 5\)
- When \(x = 2\): \(y = -50 \cdot (10)^2 = -50 \cdot 100 = -5000 \neq 5000\)
- Not a match.
Based on this analysis, the correct equation that contains both points \((-1, 5)\) and \((2, 5000)\) is:
\(y = 50 \cdot (10)^x\).