Question

Use the method of completing the square and what we know about transformations of graphs to show that the vertex of the parabola $f\left(x\right)=x^2-3x+5\mathrm{is}\mathrm{the}\mathrm{point}\left(\frac{3}{2},\frac{11}{4}\right)$

Short answer

The graph of parabola is$y=x^2$which is translating$\frac{3}{2}$units to the right and$\frac{11}{4}$units up.

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=x^2-3x+5$

Step 2. Explanation

Complete the square of the given function in order to make the function as an equation of parabola.

$$\begin{gathered} y=x^2-3x+5 \\ y={\left(x-\frac{3}{2}\right)}^2+\frac{11}{2} \end{gathered}$$

Thus, the graph of parabola is $y=x^2$which is translating $\frac{3}{2}$units to the right and $\frac{11}{4}$units up.