Write the quadratic function in the form \( y=a(x-h)^{2}+k \) and sketch its graph. \( y=x^{2}+5 x+\frac{13}{2} \)

Step 1. Starting with the quadratic function   y = x² + 5x + (13/2), we complete the square for the x‑terms. Step 2. Group the x‑terms:   x² + 5x To complete the square, take half of the coefficient of x (which is 5), so (5/2), and square it:   (5/2)² = 25/4. Step 3. Write the expression as a perfect square, subtracting the extra term:   x² + 5x = (x + 5/2)² – 25/4. Thus, substitute back into the function:   y = (x + 5/2)² – 25/4 + 13/2. Step 4. Combine the constant terms. Write 13/2 with a denominator of 4:   13/2 = 26/4. So,   –25/4 + 26/4 = 1/4. Thus, the function becomes:   y = (x + 5/2)² + 1/4. Step 5. Writing in the form y = a(x – h)² + k: Notice (x + 5/2) can be written as (x – (–5/2)). Therefore, the vertex form is:   y = 1·(x – (–5/2))² + 1/4, with   a = 1, h = –5/2, and k = 1/4. Graphing the Function: 1. Vertex: The vertex is at (h, k) = (–5/2, 1/4) or (–2.5, 0.25). 2. Orientation: Since a = 1 > 0, the parabola opens upward. 3. Axis of Symmetry: This is the vertical line x = h, so x = –2.5. 4. Y-intercept: Evaluate y at x = 0:   y = 0² + 5·0 + 13/2 = 13/2 = 6.5.   Thus, the y-intercept is at (0, 6.5). 5. X-intercepts: Solve (x + 5/2)² + 1/4 = 0:   (x + 5/2)² = –1/4. Since the square of a real number cannot be negative, there are no real x‑intercepts. To sketch: • Draw the coordinate axes. • Mark the vertex at (–2.5, 0.25). • Draw the axis of symmetry along the line x = –2.5. • Plot the y-intercept at (0, 6.5). • Since the parabola opens upward and doesn’t cross the x‑axis, sketch a smooth curve opening upward that has its minimum at the vertex. Final Answer:   The quadratic function in vertex form is   y = (x + 5/2)² + 1/4, with vertex (–5/2, 1/4) and opening upward.