Question

Let f(x) = x^2 + 16x +57.What is the vertex and minimum value of f(x)?

Answer 1

Given a quadratic function f(x):

`f(x)=ax^2+bx+c`

we can find the vertex of the parabolla that represents using the following expression:

`Vertex\colon(-(b)/(2a),f(-(b)/(2a)))`

then, in this case, we have the following values for a, b and c:

`\begin{gathered} f(x)=x^2+16x+57 \\ \text{then:} \\ a=1 \\ b=16 \\ c=57 \end{gathered}`

now lets find the x coordinate of the vertex:

`-(b)/(2a)=-(16)/(2(1))=-(16)/(2)=-8`

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):

`\begin{gathered} x=-8 \\ \Rightarrow f(-8)=(-8)^2+16(-8)+57 \\ =64-128+57=-7 \\ \Rightarrow f(-8)=-7 \end{gathered}`

we have that f(-8) = -7, therefore, the vertex of the parabola (and minimum) is located at the point (-8,-7)