https://file.aiquickdraw.com/mathbot/file/17257804626187fbgt6kz.jpg

The problem in the image reads:

"The curve with equation \( y = e^{2x} + bx + c \) passes through \( (0, 1) \) and has a stationary point at \( (2, -9) \). Find \( b \) and \( c \)."

To solve for \( b \) and \( c \), we follow these steps:

### Step 1: Use the fact that the curve passes through \( (0, 1) \).
The equation of the curve is given as \( y = e^{2x} + bx + c \). When \( x = 0 \), \( y = 1 \). Substituting \( x = 0 \) and \( y = 1 \) into the equation:

\[
1 = e^{2(0)} + b(0) + c
\]
\[
1 = 1 + c
\]
\[
c = 0
\]

### Step 2: Use the fact that the curve has a stationary point at \( (2, -9) \).
At a stationary point, the derivative \( \frac{dy}{dx} = 0 \).

First, we compute the derivative of \( y \):

\[
\frac{dy}{dx} = 2e^{2x} + b
\]

At \( x = 2 \), the slope is zero (stationary point), so we set the derivative to zero and substitute \( x = 2 \):

\[
0 = 2e^{4} + b
\]
\[
b = -2e^4
\]

### Step 3: Use the point \( (2, -9) \).
We know that at \( x = 2 \), \( y = -9 \). Substituting \( x = 2 \) and \( y = -9 \) into the original equation \( y = e^{2x} + bx + c \):

\[
-9 = e^{4} + (-2e^4)(2) + 0
\]
\[
-9 = e^4 - 4e^4
\]
\[
-9 = -3e^4
\]

Thus, the equation holds, confirming our values for \( b \) and \( c \).

### Final Answer:
\[
b = -2e^4, \quad c = 0
\]

Would you like further details or explanations on this process? Here are five related questions for further practice:

1. How do you find the stationary points of a general function?
2. What is the significance of the second derivative at a stationary point?
3. Can you explain the role of the exponential function \( e^x \) in calculus?
4. How would the solution change if the curve passed through a different point?
5. How can you verify the concavity of the curve at \( x = 2 \)?

**Tip:** When solving for constants in functions, always utilize all given conditions, such as points and derivatives, to set up multiple equations.

The curve with equation y = e^(2x) + bx + c passes through (0, 1) and has a stationary point at (2, -9). Find b and c.

This math problem involves finding the constants b and c for the curve y = e^(2x) + bx + c. The curve passes through the point (0, 1) and has a stationary point at (2, -9). Learn how to apply calculus and exponential function rules to solve this problem, suitable for Grades 11-12.

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