Find the area of the region bounded by the parabola 
y = 3x2,
 the tangent line to the parabola at (1, 3), and the x-axis.

To find the area of the region bounded by the parabola \( y = 3x^2 \), the tangent line at \( (1, 3) \), and the x-axis, follow these steps:

### Step 1: Find the equation of the tangent line.
The equation of a tangent line to a curve \( y = f(x) \) at a point \( (x_0, y_0) \) is given by:

\[
y - y_0 = f'(x_0)(x - x_0)
\]

Given the parabola \( y = 3x^2 \), its derivative is:

\[
f'(x) = \frac{d}{dx}(3x^2) = 6x
\]

At \( x = 1 \), the slope of the tangent is:

\[
f'(1) = 6(1) = 6
\]

So the tangent line at \( (1, 3) \) is:

\[
y - 3 = 6(x - 1)
\]

Simplifying this, we get the equation of the tangent line:

\[
y = 6x - 3
\]

### Step 2: Set up the integral to find the area.
The region is bounded by the parabola \( y = 3x^2 \), the tangent line \( y = 6x - 3 \), and the x-axis. We need to find the points of intersection between the parabola and the tangent line, and integrate between these points.

Set the equations equal to find the intersection points:

\[
3x^2 = 6x - 3
\]

Rearrange this into a quadratic equation:

\[
3x^2 - 6x + 3 = 0
\]

Divide by 3:

\[
x^2 - 2x + 1 = 0
\]

Factor the quadratic:

\[
(x - 1)^2 = 0
\]

Thus, \( x = 1 \). This means the parabola and the tangent intersect only at \( x = 1 \), the point of tangency.

### Step 3: Find the limits of integration.
We are also interested in where the tangent line and parabola intersect the x-axis.

- For the tangent line \( y = 6x - 3 \), set \( y = 0 \) to find the x-intercept:

\[
0 = 6x - 3 \implies x = \frac{1}{2}
\]

So the tangent line intersects the x-axis at \( x = \frac{1}{2} \).

- For the parabola \( y = 3x^2 \), set \( y = 0 \) to find the x-intercept:

\[
0 = 3x^2 \implies x = 0
\]

So the parabola intersects the x-axis at \( x = 0 \).

### Step 4: Set up the integral for the area.
The area of the region between the parabola and the tangent line from \( x = 0 \) to \( x = \frac{1}{2} \) is given by the difference between the equations of the tangent line and the parabola:

\[
\text{Area} = \int_0^{\frac{1}{2}} \left( (6x - 3) - 3x^2 \right) dx
\]

### Step 5: Compute the integral.
Now, calculate the integral:

\[
\int_0^{\frac{1}{2}} \left( 6x - 3 - 3x^2 \right) dx
\]

This simplifies to:

\[
\int_0^{\frac{1}{2}} \left( 6x - 3 - 3x^2 \right) dx = \left[ 3x^2 - 3x - x^3 \right]_0^{\frac{1}{2}}
\]

Evaluate the definite integral:

At \( x = \frac{1}{2} \):

\[
3\left( \frac{1}{2} \right)^2 - 3\left( \frac{1}{2} \right) - \left( \frac{1}{2} \right)^3 = 3\left( \frac{1}{4} \right) - 3\left( \frac{1}{2} \right) - \left( \frac{1}{8} \right) = \frac{3}{4} - \frac{3}{2} - \frac{1}{8}
\]

Simplify this expression:

\[
\frac{3}{4} - \frac{6}{4} - \frac{1}{8} = -\frac{3}{4} - \frac{1}{8} = -\frac{6}{8} - \frac{1}{8} = -\frac{7}{8}
\]

At \( x = 0 \), the integral is 0. Thus, the total area is:

\[
\text{Area} = -\left( -\frac{7}{8} \right) = \frac{7}{8}
\]

### Final Answer:
The area of the region bounded by the parabola, the tangent line, and the x-axis is \( \frac{7}{8} \) square units.

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Would you like any further details or explanations on the steps?

Here are 5 related questions:
1. How do you find the area between a curve and the x-axis in general?
2. What is the significance of the tangent line in optimization problems?
3. How do you derive the slope of a curve at a given point?
4. What is the geometric interpretation of a definite integral?
5. Can you calculate areas for bounded regions with more complex shapes?

**Tip**: When solving problems with curves and tangents, visualize the region to clearly understand the boundaries for integration.

Find the area of the region bounded by the parabola y = 3x^2, the tangent line to the parabola at (1, 3), and the x-axis.

Explore how to find the area of the region bounded by the parabola y = 3x^2, the tangent line at (1, 3), and the x-axis. This step-by-step guide uses calculus concepts like derivatives and definite integrals, suitable for Grades 11-12.

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