To find the value of $k$ and $A$, we need to use the given points and the equation of the curve, $y = kA^x$.

The points provided are:

1. (2, 50)
2. (-1, 1/5)

Step 1: Plug the first point (2, 50) into the equation

From the equation $y = kA^x$, substituting $x = 2$ and $y = 50$ gives us: $50 = kA^2$ This equation will be useful later.

Step 2: Plug the second point (-1, 1/5) into the equation

Substitute $x = -1$ and $y = \frac{1}{5}$ into the equation $y = kA^x$: $\frac{1}{5} = kA^{-1}$ This simplifies to: $\frac{1}{5} = \frac{k}{A}$ This is the second equation we will use.

Step 3: Solve the system of equations

We now have two equations:

1. $50 = kA^2$
2. $\frac{1}{5} = \frac{k}{A}$

From the second equation, we can express $k$ as: $k = \frac{A}{5}$

Step 4: Substitute $k = \frac{A}{5}$ into the first equation

Substitute this expression for $k$ into the first equation: $50 = \left( \frac{A}{5} \right) A^2$ Simplify: $50 = \frac{A^3}{5}$ Multiply both sides by 5: $250 = A^3$ Take the cube root of both sides: $A = \sqrt[3]{250}$

Step 5: Calculate $A$

Calculating the cube root of 250: $A \approx 6.3$

Step 6: Find $k$

Now that we know $A \approx 6.3$, substitute this value back into $k = \frac{A}{5}$: $k = \frac{6.3}{5} = 1.26$

Thus, the values are:

• $k \approx 1.26$
• $A \approx 6.3$

Would you like to explore more on this topic or have any further questions?

Here are 5 related questions you might find interesting:

1. How do exponential growth and decay models use similar equations?
2. What are other methods to solve for constants in equations with multiple points?
3. How does the shape of the curve change when $k$ or $A$ varies?
4. What happens when $A$ is a negative value in this type of equation?
5. How can logarithmic transformations help in solving such equations?

Tip: Always verify the units of $k$ and $A$ to ensure they match the context of the problem when dealing with real-world applications!