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We are given the function \( f \) with the conditions \( f(-4) = 5 \) and \( f'(-4) = 3 \). We are asked to find the equation of the tangent line to the function \( g(x) = \frac{f(x)}{2x+6} \) at \( x = -4 \).

### Step 1: Find the value of \( g(-4) \)
We use the given information for \( f(-4) \) to find \( g(-4) \):
\[
g(-4) = \frac{f(-4)}{2(-4) + 6} = \frac{5}{-8 + 6} = \frac{5}{-2} = -\frac{5}{2}
\]

### Step 2: Find the derivative of \( g(x) \)
To find the slope of the tangent line, we need to calculate \( g'(x) \). We will use the quotient rule for differentiation, which states:
\[
\left( \frac{f(x)}{h(x)} \right)' = \frac{f'(x)h(x) - f(x)h'(x)}{h(x)^2}
\]
In our case, \( h(x) = 2x + 6 \), so \( h'(x) = 2 \). Applying the quotient rule, we get:
\[
g'(x) = \frac{f'(x)(2x + 6) - f(x)(2)}{(2x + 6)^2}
\]

### Step 3: Evaluate \( g'(-4) \)
Substitute \( x = -4 \), \( f(-4) = 5 \), and \( f'(-4) = 3 \) into the expression for \( g'(x) \):
\[
g'(-4) = \frac{3(-8 + 6) - 5(2)}{(-8 + 6)^2} = \frac{3(-2) - 10}{(-2)^2} = \frac{-6 - 10}{4} = \frac{-16}{4} = -4
\]

### Step 4: Write the equation of the tangent line
The equation of a line in point-slope form is:
\[
y - y_1 = m(x - x_1)
\]
Here, \( x_1 = -4 \), \( y_1 = g(-4) = -\frac{5}{2} \), and the slope \( m = g'(-4) = -4 \). Substituting these values, we get:
\[
y - \left(-\frac{5}{2}\right) = -4(x + 4)
\]
Simplifying:
\[
y + \frac{5}{2} = -4(x + 4)
\]
\[
y = -4(x + 4) - \frac{5}{2}
\]
\[
y = -4x - 16 - \frac{5}{2}
\]
Converting \( -16 \) to a fraction with denominator 2:
\[
y = -4x - \frac{32}{2} - \frac{5}{2}
\]
\[
y = -4x - \frac{37}{2}
\]
Thus, the equation of the tangent line is:
\[
y = -4x - \frac{37}{2}
\]

Would you like more details or have any questions?

Here are 5 related questions to extend your understanding:
1. How would you apply the quotient rule to a more complex function?
2. What if \( f'(x) \) were a more complicated expression, how would that affect the derivative of \( g(x) \)?
3. How can we interpret the slope of the tangent line geometrically?
4. Can you explain why the quotient rule works in terms of the product and chain rules?
5. How would the tangent line equation change if the point of tangency were at a different \( x \)-value?

**Tip:** When applying the quotient rule, always differentiate the numerator and denominator carefully to avoid errors in signs.

Let f be a differentiable function with f(-4) = 5 and f'(-4) = 3. Let the function g(x) = f(x) / (2x + 6). Write the equation of the line tangent to the graph of g at the point where x = -4.

Learn how to calculate the tangent line to the graph of g(x) = f(x)/(2x+6) at x = -4 using the quotient rule. The given function has f(-4) = 5 and f'(-4) = 3. This solution includes differentiation and equation of the tangent line.

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