To find the equation of the tangent line to the function $f(x) = 2x^2 + 4$ at $x = 1$, we can follow these steps:

Step 1: Find the derivative of the function

The derivative $f'(x)$ will give us the slope of the tangent line at any point $x$.

$f'(x) = \frac{d}{dx}(2x^2 + 4) = 4x$

Step 2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into the derivative to find the slope of the tangent line at that point.

$f'(1) = 4(1) = 4$

Thus, the slope of the tangent line at $x = 1$ is 4.

Step 3: Find the point on the curve at $x = 1$

To find the y-coordinate at $x = 1$, substitute $x = 1$ into the original function $f(x)$.

$f(1) = 2(1)^2 + 4 = 2 + 4 = 6$

Thus, the point of tangency is $(1, 6)$.

Step 4: Use the point-slope form to write the equation of the tangent line

The point-slope form of the equation of a line is:

$y - y_1 = m(x - x_1)$

Where:

• $m$ is the slope (which is 4),
• $(x_1, y_1)$ is the point of tangency $(1, 6)$.

Substitute these values into the equation:

$y - 6 = 4(x - 1)$

Simplifying:

$y - 6 = 4x - 4$ $y = 4x - 4 + 6$ $y = 4x + 2$

Final Answer:

The equation of the tangent line is $y = 4x + 2$.

Would you like any further details on how this was derived? Here are some related questions you can consider:

1. What are other methods to find the equation of a tangent line?
2. How can you verify that the tangent line and the curve intersect at only one point?
3. How does the derivative of a function geometrically represent the slope of the tangent line?
4. What happens to the tangent line when the function changes, for example, if $f(x)$ is a cubic function?
5. Can you find the tangent line to this function at other points, like $x = 2$?

Tip:

Always remember to double-check both the derivative and the point of tangency when solving tangent line problems to avoid small errors.