Question

How do you find the vertex of a quadratic function?f(x) = 3x^2+18x+32

Answer 1

Given the following quadratic function:

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

We can find the vertex with the following general rule:

`V=(-(b)/(2a),f(-(b)/(2a)))`

in this case, we have the following quadratic function:

`f(x)=3x^2+18x+32`

then, the coefficients a,b and c are:

`\begin{gathered} a=3 \\ b=18 \\ c=32 \end{gathered}`

now, we can calculate first the x coordinate of the vertex:

`-(b)/(2a)=-(18)/(2(3))=-(18)/(6)=-3`

then, we have to evaluate x = -3 on the function to get the y coordinate:

`\begin{gathered} f(-3)=3(-3)^2+18(-3)+32 \\ =3(9)-54+32=27-54+32=5 \\ f(-3)=5 \end{gathered}`

therefore, the vertex of the function f(x) is:

`V=(-3,f(-3))=(-3,5)`