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The image presents a calculus problem involving finding the point on the graph of the function \( y = \frac{1}{2}x^2 \) where the tangent line is parallel to a given line, and then determining the equation of that tangent line. Here's a detailed explanation of the solution process:

### Problem:
Given the graph of \( y = \frac{1}{2}x^2 \), find the point on the graph where the tangent line is parallel to the line \( 2x - 4y = 3 \). Then, write the equation of this tangent line.

### Solution:

1. **Find the Slope of the Given Line:**
   The given line is \( 2x - 4y = 3 \). To find its slope, rewrite this equation in slope-intercept form \( y = mx + b \).

   \[
   2x - 4y = 3 \quad \Rightarrow \quad -4y = -2x + 3 \quad \Rightarrow \quad y = \frac{1}{2}x - \frac{3}{4}
   \]

   The slope \( m \) of this line is \( \frac{1}{2} \).

2. **Find the Derivative of the Given Function:**
   The function is \( y = \frac{1}{2}x^2 \). To find the slope of the tangent line at any point on this curve, take the derivative of \( y \) with respect to \( x \):

   \[
   \frac{dy}{dx} = y' = x
   \]

   The derivative \( y' = x \) gives the slope of the tangent line at any point \( x \) on the curve.

3. **Set the Derivative Equal to the Slope of the Given Line:**
   For the tangent line to be parallel to the given line, its slope must equal \( \frac{1}{2} \):

   \[
   x = \frac{1}{2}
   \]

   This value of \( x \) is the point where the tangent line is parallel to the given line.

4. **Find the Corresponding \( y \)-Value:**
   Substitute \( x = \frac{1}{2} \) into the original equation \( y = \frac{1}{2}x^2 \) to find the corresponding \( y \)-value:

   \[
   y = \frac{1}{2} \left(\frac{1}{2}\right)^2 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}
   \]

   So, the point on the graph is \( \left(\frac{1}{2}, \frac{1}{8}\right) \).

5. **Write the Equation of the Tangent Line:**
   The tangent line at this point has the slope \( \frac{1}{2} \) and passes through \( \left(\frac{1}{2}, \frac{1}{8}\right) \). Use the point-slope form of the equation of a line:

   \[
   y - y_1 = m(x - x_1)
   \]

   Substitute \( m = \frac{1}{2} \), \( x_1 = \frac{1}{2} \), and \( y_1 = \frac{1}{8} \):

   \[
   y - \frac{1}{8} = \frac{1}{2}\left(x - \frac{1}{2}\right)
   \]

   Simplify to get the equation of the tangent line:

   \[
   y = \frac{1}{2}x - \frac{1}{4} + \frac{1}{8} = \frac{1}{2}x - \frac{1}{8}
   \]

### Final Answer:
The equation of the tangent line parallel to the given line is:

\[
y = \frac{1}{2}x - \frac{1}{8}
\]

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Would you like further details or have any questions? Here are some related questions to explore further:

1. How do you determine the slope of a line given in standard form?
2. What is the significance of the derivative in finding tangent lines?
3. How do you apply the point-slope form to find the equation of a line?
4. How do you determine if two lines are parallel?
5. What is the geometric interpretation of the derivative at a point on a curve?

**Tip:** Always express the line equation in slope-intercept form to easily identify the slope and y-intercept.

Given the graph of y = (1/2)x^2, at what point on the graph is the tangent line parallel to the line 2x - 4y = 3? Write an equation of this tangent line.

This problem involves finding the tangent line to the graph of y = (1/2)x^2 that is parallel to the line 2x - 4y = 3. Learn how to determine the point of tangency and write the equation of the tangent line using calculus concepts like differentiation and the slope of a line. Suitable for students in Grades 10-12.

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