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The question involves the curve \( y = 2e^{2x} + e^{-x} \) and asks to find the equation of the tangent at a point \( P \), where this tangent cuts the x-axis. Let's break down the problem step by step.

### 1. Find the point where the curve cuts the y-axis.
To find the y-intercept (where the curve cuts the y-axis), we set \( x = 0 \) in the equation:

\[
y = 2e^{2(0)} + e^{-(0)} = 2e^0 + e^0 = 2 + 1 = 3
\]

So the point where the curve cuts the y-axis is \( (0, 3) \).

### 2. Find the slope of the tangent line at the point.
To find the equation of the tangent at any point, we need the derivative of \( y \). Let's differentiate \( y = 2e^{2x} + e^{-x} \):

\[
\frac{dy}{dx} = 2 \cdot 2e^{2x} - e^{-x} = 4e^{2x} - e^{-x}
\]

At the point \( (0, 3) \), substitute \( x = 0 \) into the derivative to find the slope of the tangent:

\[
\frac{dy}{dx} \bigg|_{x=0} = 4e^{2(0)} - e^{-(0)} = 4e^0 - e^0 = 4 - 1 = 3
\]

So, the slope of the tangent at the point \( (0, 3) \) is 3.

### 3. Find the equation of the tangent.
The equation of the tangent line is given by the point-slope form:

\[
y - y_1 = m(x - x_1)
\]

Where \( (x_1, y_1) \) is the point of tangency, and \( m \) is the slope. Substituting the values \( x_1 = 0 \), \( y_1 = 3 \), and \( m = 3 \):

\[
y - 3 = 3(x - 0)
\]

Simplifying:

\[
y = 3x + 3
\]

So the equation of the tangent is \( y = 3x + 3 \).

### 4. Find the x-intercept of the tangent line.
To find where the tangent cuts the x-axis, set \( y = 0 \) in the tangent equation:

\[
0 = 3x + 3
\]

Solving for \( x \):

\[
3x = -3 \quad \Rightarrow \quad x = -1
\]

Thus, the tangent cuts the x-axis at \( x = -1 \), so the point of intersection is \( (-1, 0) \).

### Final Answer:
The equation of the tangent to the curve at the point \( (0, 3) \) is \( y = 3x + 3 \), and the tangent cuts the x-axis at the point \( (-1, 0) \).

Let me know if you'd like further details or if you have any questions!

### Additional Related Questions:
1. How do you find the second derivative of the curve?
2. How would the equation of the tangent change if the curve was shifted vertically?
3. What is the nature of the curve, given its exponential terms?
4. How would you find the equation of a normal to the curve at a given point?
5. How would you solve for tangents at other points along the curve?

### Tip:
When differentiating exponential functions, remember the chain rule. For instance, \( \frac{d}{dx}e^{2x} = 2e^{2x} \).

The curve y = 2e^(2x) + e^(-x) cuts the y-axis at the point P. Find the coordinates of the point where the curve intersects the y-axis and the equation of the tangent at the point where this tangent cuts the x-axis.

Learn how to find the equation of the tangent line for the curve y = 2e^(2x) + e^(-x) and determine where it intersects the axes. This problem involves exponential functions, differentiation, and the point-slope form of a line. Suitable for students in Grades 11-12.

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