The task is to find the equation of the tangent line to the graph of the function $f(x) = \sin\left( 2x + \frac{\pi}{6} \right)$ at the point where the graph intersects the y-axis.

Step 1: Find the point of intersection with the y-axis

To find where the graph intersects the y-axis, set $x = 0$, since at the y-axis, $x = 0$.

$f(0) = \sin\left( 2(0) + \frac{\pi}{6} \right) = \sin\left( \frac{\pi}{6} \right) = \frac{1}{2}$

Thus, the point of intersection is $(0, \frac{1}{2})$.

Step 2: Find the derivative of the function

The slope of the tangent line is given by the derivative of $f(x)$ at the point $x = 0$.

We differentiate $f(x) = \sin\left( 2x + \frac{\pi}{6} \right)$ using the chain rule.

$f'(x) = \cos\left( 2x + \frac{\pi}{6} \right) \cdot 2$

Now, evaluate the derivative at $x = 0$:

$f'(0) = 2 \cdot \cos\left( 2(0) + \frac{\pi}{6} \right) = 2 \cdot \cos\left( \frac{\pi}{6} \right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

Thus, the slope of the tangent line at $x = 0$ is $\sqrt{3}$.

Step 3: Equation of the tangent line

The equation of the tangent line at a point $(x_0, y_0)$ with slope $m$ is:

$y - y_0 = m(x - x_0)$

Substitute $x_0 = 0$, $y_0 = \frac{1}{2}$, and $m = \sqrt{3}$:

$y - \frac{1}{2} = \sqrt{3}(x - 0)$

Simplifying this:

$y = \sqrt{3}x + \frac{1}{2}$

Thus, the equation of the tangent line is:

$y = \sqrt{3}x + \frac{1}{2}$

Would you like further clarification on any steps?

Related Questions:

1. How do you find the derivative of trigonometric functions in general?
2. What is the significance of the slope in the context of tangent lines?
3. Can the method used here be applied to any trigonometric function?
4. How would the equation change if the function was $\cos(2x + \frac{\pi}{6})$ instead?
5. How do you find points of intersection between a curve and any arbitrary axis?

Tip:

When finding tangent lines, always start by identifying the point of interest and computing the derivative at that point, as it gives you the slope of the tangent.