Find the equation of the line tangent to the graph of g(x)= 2x^2 -3 over 3x+1 when x ==1.

To find the equation of the tangent line to the graph of \( g(x) = \frac{2x^2 - 3}{3x + 1} \) at \( x = 1 \), we need two things:

1. **The point of tangency** (i.e., \( (1, g(1)) \)).
2. **The slope of the tangent line** at \( x = 1 \), which we find by differentiating \( g(x) \) and evaluating \( g'(1) \).

### Step 1: Calculate \( g(1) \)
Substitute \( x = 1 \) into \( g(x) \):
\[
g(1) = \frac{2(1)^2 - 3}{3(1) + 1} = \frac{2 - 3}{3 + 1} = \frac{-1}{4} = -\frac{1}{4}.
\]
So, the point of tangency is \( (1, -\frac{1}{4}) \).

### Step 2: Find \( g'(x) \) using the Quotient Rule
To find \( g'(x) \), we use the Quotient Rule:
\[
g'(x) = \frac{(3x + 1) \cdot \frac{d}{dx}(2x^2 - 3) - (2x^2 - 3) \cdot \frac{d}{dx}(3x + 1)}{(3x + 1)^2}.
\]
Calculate each derivative:

- The derivative of the numerator, \( 2x^2 - 3 \), is \( 4x \).
- The derivative of the denominator, \( 3x + 1 \), is \( 3 \).

Thus,
\[
g'(x) = \frac{(3x + 1)(4x) - (2x^2 - 3)(3)}{(3x + 1)^2}.
\]
Expanding and simplifying the numerator:
\[
= \frac{12x^2 + 4x - (6x^2 - 9)}{(3x + 1)^2} = \frac{6x^2 + 4x + 9}{(3x + 1)^2}.
\]

### Step 3: Evaluate \( g'(1) \)
Now, substitute \( x = 1 \) into \( g'(x) \):
\[
g'(1) = \frac{6(1)^2 + 4(1) + 9}{(3(1) + 1)^2} = \frac{6 + 4 + 9}{4^2} = \frac{19}{16}.
\]
So, the slope of the tangent line at \( x = 1 \) is \( \frac{19}{16} \).

### Step 4: Write the Equation of the Tangent Line
Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), with \( (x_1, y_1) = (1, -\frac{1}{4}) \) and \( m = \frac{19}{16} \), we get:
\[
y + \frac{1}{4} = \frac{19}{16}(x - 1).
\]
Expanding and rearranging, we have:
\[
y = \frac{19}{16}x - \frac{19}{16} - \frac{1}{4} = \frac{19}{16}x - \frac{23}{16}.
\]

### Final Answer
The equation of the tangent line is:
\[
y = \frac{19}{16}x - \frac{23}{16}.
\]

Would you like a breakdown of any steps, or do you have any other questions?

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### Related Questions
1. How do you apply the Quotient Rule in other contexts?
2. How would you find the normal line to the curve at \( x = 1 \)?
3. How does the slope of the tangent line change as \( x \) varies?
4. What are alternative methods for finding the slope at a point on a function?
5. How could you determine the concavity of \( g(x) \) at \( x = 1 \)?

### Tip
Always confirm if simplifying the expression of derivatives or tangent lines makes future calculations easier, especially with complex functions.

Discover how to determine the equation of the tangent line to g(x) = (2x^2 - 3) / (3x + 1) at x = 1. This solution demonstrates the use of differentiation, specifically the Quotient Rule, to calculate the slope at a point and find the tangent line equation. Ideal for students in Grades 11-12.

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