We know that a parabolla is given by the form:
![p(x)=ax^2+bx+c](https://img.qammunity.org/2023/formulas/mathematics/college/l1mewrtssrl8w2f5zd2bc4wvssjjo3y4vm.png)
We also know that we can factor this equation to get the form:
![p(x)=(x-h)(x-k)](https://img.qammunity.org/2023/formulas/mathematics/college/cbbr6g6x32fsay80fjh5tnxdulud2hphnk.png)
Where
![p(h)=0\text{ and }p(k)=0](https://img.qammunity.org/2023/formulas/mathematics/college/ln16o68y3sd1hm195gy49r8mae68tldruq.png)
This is, h and k are the roots.
Then we can write:
h = 14
k = 11
![p(x)=(x-14)(x-11)](https://img.qammunity.org/2023/formulas/mathematics/college/5zgetvfc26paojr6jnqi6qvwt6r443kbr9.png)
We can also doing the multiplication to give the answer is standard form a parabolla:
![\begin{gathered} p(x)=(x-14)(x-11)=x^2-11x-14x+154 \\ p(x)=x^2-25x+154 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w9qs63gvm962i791tsh8kyg3w026yjghwy.png)
And that's a function as aked in the problem.