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

Function by following steps: - step0: Rewrite in vertex form: \(y=x^{2}+2x+\frac{5}{2}\) - step1: Group the terms: \(y=\left(x^{2}+2x\right)+\frac{5}{2}\) - step2: Add and subtract: \(y=\left(x^{2}+2x+1-1\right)+\frac{5}{2}\) - step3: Use the distributive property: \(y=\left(x^{2}+2x+1\right)-1+\frac{5}{2}\) - step4: Factor the expression: \(y=\left(x+1\right)^{2}-1+\frac{5}{2}\) - step5: Calculate: \(y=\left(x+1\right)^{2}+\frac{3}{2}\) I apologize for the inconvenience; it seems that I cannot generate a plot at the moment. However, I can describe how to sketch the graph based on the information we have: 1. **Vertex**: Plot the vertex at the point \( (-1, \frac{3}{2}) \). 2. **Y-Intercept**: Plot the y-intercept at \( (0, \frac{5}{2}) \). 3. **Direction**: Since the parabola opens upwards, draw a U-shaped curve starting from the vertex and passing through the y-intercept. 4. **Symmetry**: The parabola is symmetric about the vertical line \( x = -1 \). You can find points on either side of the vertex to help shape the curve. ### Key Points to Plot: - Vertex: \( (-1, \frac{3}{2}) \) - Y-Intercept: \( (0, \frac{5}{2}) \) - Since there are no real x-intercepts, the graph will not cross the x-axis. This should give you a clear idea of how the graph looks. If you have graphing software or graph paper, you can use these points to create an accurate sketch.