To determine the point(s) where the graph of the function has a horizontal tangent line, we need to find the points where the derivative of the function equals zero.

Given the function:

$y = \frac{1}{2} x^2 + 7x$

Step 1: Find the derivative of the function.

To find the slope of the tangent line at any point, we compute the derivative of $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2} x^2 + 7x \right)$

Using basic differentiation rules:

$\frac{dy}{dx} = \frac{1}{2} \cdot 2x + 7 = x + 7$

Step 2: Set the derivative equal to zero.

A horizontal tangent line occurs where the slope of the curve is zero. Therefore, set the derivative equal to zero:

$x + 7 = 0$

Solving for $x$:

$x = -7$

Step 3: Find the corresponding $y$-coordinate.

Substitute $x = -7$ back into the original equation to find the corresponding $y$-value:

$y = \frac{1}{2} (-7)^2 + 7(-7) = \frac{1}{2} \times 49 - 49 = 24.5 - 49 = -24.5$

Conclusion:

The graph has a horizontal tangent line at the point $(-7, -24.5)$.

Would you like a more detailed explanation or have any questions about the solution? Here are some related questions:

1. How do you differentiate more complex functions involving higher powers of $x$?
2. What are some applications of finding the slope of a tangent line in real-world problems?
3. How do you determine if a function has an inflection point or a local extremum?
4. What are the geometric interpretations of the derivative at various points on a curve?
5. How would you find horizontal tangents for parametric or implicit functions?

Tip: When finding horizontal tangents, always ensure you're solving for where the derivative equals zero, as this represents zero slope!