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:

$V=(-\frac{b}{2a},f(-\frac{b}{2a}))$

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:

$-\frac{b}{2a}=-\frac{18}{2(3)}=-\frac{18}{6}=-3$

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:

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