find an equation of the tangent line to the graph of the function y=(2x-3)/(2x+3) at the point (3,1/3)

To find the equation of the tangent line to the graph of the function \( y = \frac{2x - 3}{2x + 3} \) at the point \( (3, \frac{1}{3}) \), we follow these steps:

### Step 1: Differentiate the function
We need to find the derivative of the function \( y \) with respect to \( x \) (denoted as \( \frac{dy}{dx} \)), because the derivative at a point gives us the slope of the tangent line at that point.

Given:
\[
y = \frac{2x - 3}{2x + 3}
\]
We will apply the **quotient rule** for differentiation. The quotient rule states:
\[
\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}
\]
Here, \( f(x) = 2x - 3 \) and \( g(x) = 2x + 3 \). Now, differentiate both functions:
- \( f'(x) = 2 \)
- \( g'(x) = 2 \)

Using the quotient rule:
\[
\frac{dy}{dx} = \frac{(2)(2x + 3) - (2x - 3)(2)}{(2x + 3)^2}
\]
Simplifying the numerator:
\[
\frac{dy}{dx} = \frac{4x + 6 - (4x - 6)}{(2x + 3)^2}
\]
\[
\frac{dy}{dx} = \frac{4x + 6 - 4x + 6}{(2x + 3)^2}
\]
\[
\frac{dy}{dx} = \frac{12}{(2x + 3)^2}
\]

### Step 2: Find the slope at \( x = 3 \)
Now, substitute \( x = 3 \) into the derivative to find the slope at the point \( (3, \frac{1}{3}) \):
\[
\frac{dy}{dx} \bigg|_{x=3} = \frac{12}{(2(3) + 3)^2} = \frac{12}{(6 + 3)^2} = \frac{12}{9^2} = \frac{12}{81} = \frac{4}{27}
\]
So, the slope of the tangent line at \( x = 3 \) is \( \frac{4}{27} \).

### Step 3: Use the point-slope form of the line
The equation of the tangent line can be found using the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
Where \( (x_1, y_1) = (3, \frac{1}{3}) \) and \( m = \frac{4}{27} \). Substituting these values:
\[
y - \frac{1}{3} = \frac{4}{27}(x - 3)
\]
Simplifying:
\[
y - \frac{1}{3} = \frac{4}{27}x - \frac{12}{27}
\]
\[
y - \frac{1}{3} = \frac{4}{27}x - \frac{4}{9}
\]
Adding \( \frac{1}{3} \) to both sides:
\[
y = \frac{4}{27}x - \frac{4}{9} + \frac{1}{3}
\]
Convert \( \frac{1}{3} \) to a denominator of 9:
\[
y = \frac{4}{27}x - \frac{4}{9} + \frac{3}{9}
\]
\[
y = \frac{4}{27}x - \frac{1}{9}
\]

Thus, the equation of the tangent line is:
\[
y = \frac{4}{27}x - \frac{1}{9}
\]

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

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Here are 5 related questions:
1. How do you differentiate a quotient of two functions?
2. How can you verify that the point \( (3, \frac{1}{3}) \) lies on the given curve?
3. What happens to the slope of a tangent line at points where the function is undefined?
4. How would the tangent line equation change if the point were different?
5. Can you derive the equation of a normal line at the same point?

**Tip:** The quotient rule is useful when differentiating ratios of functions. Practice applying it carefully to avoid mistakes with signs!

Find the equation of the tangent line to the function y=(2x-3)/(2x+3) at the point (3, 1/3). This problem covers key calculus concepts such as differentiation using the quotient rule and finding the slope of a tangent line.

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