For f(x) = 1/2+x^2 the slope of the graph of y = f(x) is known to be -2/9 at the point with x-coordinate 1. Find the equation of the tangent line at that point.

To find the equation of the tangent line to the graph of \( y = f(x) \) at the point where \( x = 1 \), we'll follow these steps:

### Step 1: Recall the equation of the tangent line
The equation of the tangent line at a point \( x = a \) is given by the formula:
\[
y - y_1 = m(x - x_1)
\]
where:
- \( (x_1, y_1) \) is the point of tangency,
- \( m \) is the slope of the tangent line.

In this case, we are given:
- The slope of the tangent line, \( m = -\frac{2}{9} \),
- The \( x \)-coordinate of the point of tangency, \( x_1 = 1 \).

We still need to determine \( y_1 \), the corresponding \( y \)-coordinate of the point on the curve at \( x = 1 \).

### Step 2: Find the \( y \)-coordinate of the point
The \( y \)-coordinate is found by evaluating \( f(x) \) at \( x = 1 \). The function is:
\[
f(x) = \frac{1}{2 + x^2}
\]
Substitute \( x = 1 \) into the function:
\[
f(1) = \frac{1}{2 + 1^2} = \frac{1}{2 + 1} = \frac{1}{3}
\]
Thus, \( y_1 = \frac{1}{3} \).

### Step 3: Write the equation of the tangent line
Now, we have all the necessary information:
- \( m = -\frac{2}{9} \),
- \( (x_1, y_1) = (1, \frac{1}{3}) \).

Substitute these values into the point-slope form of the line equation:
\[
y - \frac{1}{3} = -\frac{2}{9}(x - 1)
\]

### Step 4: Simplify the equation
Distribute the slope on the right-hand side:
\[
y - \frac{1}{3} = -\frac{2}{9}x + \frac{2}{9}
\]
Now, add \( \frac{1}{3} \) to both sides to isolate \( y \):
\[
y = -\frac{2}{9}x + \frac{2}{9} + \frac{1}{3}
\]
Convert \( \frac{1}{3} \) to have a denominator of 9:
\[
\frac{1}{3} = \frac{3}{9}
\]
Thus, the equation becomes:
\[
y = -\frac{2}{9}x + \frac{2}{9} + \frac{3}{9} = -\frac{2}{9}x + \frac{5}{9}
\]

### Final Answer: 
The equation of the tangent line is:
\[
y = -\frac{2}{9}x + \frac{5}{9}
\]

Would you like more details or have any questions about this process?

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Here are 5 related questions to expand your understanding:
1. How do you find the slope of the tangent line for a given function if it's not provided?
2. What role does the derivative play in finding the slope of a tangent line?
3. Can the tangent line to a curve ever intersect the curve at more than one point?
4. What is the difference between a secant line and a tangent line to a curve?
5. How can the equation of a normal line to a curve be derived from the tangent line?

**Tip:** The slope of a tangent line at any point is equal to the value of the derivative of the function at that point.

Find the equation of the tangent line to the curve f(x) = 1/(2+x^2) at x = 1, where the slope is -2/9. This detailed solution walks through point-slope form and function evaluation. Ideal for calculus students in Grades 11-12.

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