Given a quadratic function f(x):
[tex]f(x)=ax^2+bx+c[/tex]
we can find the vertex of the parabolla that represents using the following expression:
[tex]Vertex\colon(-\frac{b}{2a},f(-\frac{b}{2a}))[/tex]
then, in this case, we have the following values for a, b and c:
[tex]\begin{gathered} f(x)=x^2+16x+57 \\ \text{then:} \\ a=1 \\ b=16 \\ c=57 \end{gathered}[/tex]
now lets find the x coordinate of the vertex:
[tex]-\frac{b}{2a}=-\frac{16}{2(1)}=-\frac{16}{2}=-8[/tex]
next, to find the y-coordinate of the vertex, we have to evaluate the x-coordinate on the function, that is, we have to find f(-8):
[tex]\begin{gathered} x=-8 \\ \Rightarrow f(-8)=(-8)^2+16(-8)+57 \\ =64-128+57=-7 \\ \Rightarrow f(-8)=-7 \end{gathered}[/tex]
we have that f(-8) = -7, therefore, the vertex of the parabola (and minimum) is located at the point (-8,-7)
