Let's suppose that we have the following function in standard form:
![f(x)=(x-1)^2-4](https://img.qammunity.org/2023/formulas/mathematics/college/rem3apx524fba0nhgq0b5di6ewwwlkb6gp.png)
We know that the general standard form of a function is:
![f(x)=a(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/high-school/lv6wh92oxxg1yzd73cyhfmkxhau9bpvca1.png)
then, in this case we have the following:
![\begin{gathered} a=1 \\ (h,k)=(1,-4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x366v139uabjdypvrnh1wmmf2qhiddrg6i.png)
Since the function is quadratic, the graph will be a parabola. The point (h,k) is where the vertex is located on the plane.
Then, we can find the roots of the function to see where the intersections with the x-axis are (if there are any):
![\begin{gathered} f(x)=(x-1)^2-4=0 \\ \Rightarrow(x-1)^2-4=0 \\ \Rightarrow x^2-2x+1-4=0 \\ \Rightarrow x^2-2x-3=0 \\ \Rightarrow(x+1)(x-3)=0 \\ \Rightarrow x_1=-1 \\ \text{and} \\ x_2=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wequ64qae8c5pp0x12pay0uvy9irb9ldt2.png)
we have that the parabola will cross the x axis when x=-1 and x=3, then, a sketch of the graph of f(x) would look like this: