The given function is
![f(x)=x^2+5x-2](https://img.qammunity.org/2023/formulas/mathematics/college/yfz4emmj5kjko6k3hrx03zvw4bki9i5tup.png)
The vertex has two coordinates, h, and k. To find h we use the following formula.
![h=-(b)/(2a)](https://img.qammunity.org/2023/formulas/mathematics/college/cwwf9bbin9d0ejklymkv2ou3q3tng71xg4.png)
Where a = 1 and b = 5.
![h=-(5)/(2(1))=-(5)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/iy6705sm6swjf22pi45w9ke5r9nw09lj40.png)
Then, we evaluate the function when x = -5/2 to find k.
![\begin{gathered} k=f(-(5)/(2))=(-(5)/(2))^2+5(-(5)/(2))-2 \\ k=(25)/(4)-(25)/(2)-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ydt13zc0o6q8m7dpae1u5gkwnh0ni6pf5f.png)
Let's find the least common factor between the denominators.
4 2 | 2
2 1 | 2
1
The least common factor would be 2*2 = 4, let's use it.
![\begin{gathered} k=(25-2\cdot25-4\cdot2)/(4) \\ k=(25-50-8)/(4) \\ k=-(33)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/viai3snxap7bzzpymf91k8tsm41avl9vw1.png)
Once we know the value of h and k, we can deduct the vertex.
Therefore, the vertex is (-5/2, -33/4).
The standard form of a quadratic function is
![f(x)=a(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/high-school/lv6wh92oxxg1yzd73cyhfmkxhau9bpvca1.png)
In this case, a = 1, h = -5/2, and k = -33/4.
So, the standard form of f(x) is
![f(x)=(x+(5)/(2))^2-(33)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/gbokflffzvp4l0qqiosb479udvf4cvi1py.png)