Find the minimum value of the parabola y=x^2-3x Simplify your answer and write it as a proper fraction improper fraction, or integer. square

To find the minimum value of the parabola $y=x^{2}-3x$, we can complete the square to equation in vertex form.<br /><br />Step 1: Start with the given equation:<br />$y=x^{2}-3x$<br /><br />Step 2: Complete the square by adding and subtracting $\left(\frac{3}{2}\right)^{2}$ inside the parentheses:<br />$y=\left(x^{2}-3x+\left(\frac{3}{2}\right)^{2}\right)-\left(\frac{3}{2}\right)^{2}$<br />$y=\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}$<br /><br />Step 3: The vertex form of a parabola is $y=a(x-h)^{2}+k$, where $(h,k)$ is the vertex of the parabola. In this case, the vertex is $\left(\frac{3}{2},-\frac{9}{4}\right)$.<br /><br />Step 4: Since the coefficient of $(x-\frac{3}{2})^{2}$ is positive, the parabola opens upwards. Therefore, the vertex represents the minimum point of the parabola.<br /><br />Step 5: The minimum value of the parabola is the y-coordinate of the vertex, which is $-\frac{9}{4}$.<br /><br />Therefore, the minimum value of the parabola $y=x^{2}-3x$ is $-\frac{9}{4}$.