Question

find the vertex the line of symmetry the maximum or minimum value of the quadratic function f(x)=3x^2-18+13

Answer 1

vertex at
\[\frac{b}{2a}=-\frac{-18}{2\times 3}=3\] for the first coordinate. find the second by replacing x by 3 to see what you get

Answer 2

i am assuming you meant
\[f(x)=3x^2-18x+13\]

Answer 3

you should get
\[f(3)=-14\] so vertex is (3,-14) and that is a minimum since parabola opens up (leading coefficient is positive)