We need to find the area of the region bounded by the parabola $y = 2x^2$, the tangent line to this parabola at the point $(2, 8)$, and the x-axis.

Step 1: Equation of the tangent line

To find the equation of the tangent line at the point $(2, 8)$, we need the slope of the parabola at $x = 2$. The slope of the tangent line is given by the derivative of $y = 2x^2$.

$\frac{dy}{dx} = \frac{d}{dx}(2x^2) = 4x$

At $x = 2$:

$\frac{dy}{dx} = 4(2) = 8$

So the slope of the tangent line is 8. Using the point-slope form of the equation of a line $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1) = (2, 8)$, the equation of the tangent line is:

$y - 8 = 8(x - 2)$

Simplifying:

$y - 8 = 8x - 16$ $y = 8x - 8$

Thus, the equation of the tangent line is $y = 8x - 8$.

Step 2: Points of intersection between the parabola and the tangent line

We need to find the points where the parabola $y = 2x^2$ and the tangent line $y = 8x - 8$ intersect.

Set the two equations equal to each other:

$2x^2 = 8x - 8$ $2x^2 - 8x + 8 = 0$ $x^2 - 4x + 4 = 0$ $(x - 2)^2 = 0$ $x = 2$

The parabola and the tangent line intersect only at $x = 2$.

Step 3: Points of intersection with the x-axis

To find where the parabola and tangent line intersect the x-axis, set $y = 0$ in both equations.

For the parabola $y = 2x^2$:

$2x^2 = 0 \quad \Rightarrow \quad x = 0$

So, the parabola intersects the x-axis at $(0, 0)$.

For the tangent line $y = 8x - 8$:

$8x - 8 = 0 \quad \Rightarrow \quad x = 1$

So, the tangent line intersects the x-axis at $(1, 0)$.

Step 4: Area of the region

The region is bounded by the parabola, the tangent line, and the x-axis between $x = 0$ and $x = 1$. We will compute the area as the integral of the difference between the parabola and the tangent line.

The area $A$ is given by:

$A = \int_0^1 \left( (8x - 8) - 2x^2 \right) dx$

Simplifying the integrand:

$A = \int_0^1 (8x - 8 - 2x^2) dx = \int_0^1 (-2x^2 + 8x - 8) dx$

Now, integrate each term:

$A = \left[ -\frac{2x^3}{3} + 4x^2 - 8x \right]_0^1$

Evaluating at $x = 1$:

$A = \left( -\frac{2(1)^3}{3} + 4(1)^2 - 8(1) \right) - \left( 0 \right)$ $A = \left( -\frac{2}{3} + 4 - 8 \right)$ $A = -\frac{2}{3} - 4$ $A = -\frac{14}{3}$

Since the area cannot be negative, we take the absolute value:

$A = \frac{14}{3}$

Thus, the area of the region is $\frac{14}{3}$ square units.

Would you like more details or clarifications?

Here are 5 related questions to deepen your understanding:

1. How do you find the equation of a tangent line to any curve?
2. What is the geometric significance of finding the area between curves?
3. Can this method be applied to any two intersecting curves?
4. What are other ways to find points of intersection between two curves?
5. How would the area change if the parabola had a different coefficient?

Tip: Always remember to check for symmetry when calculating areas. This can sometimes simplify the work.