Find the vertex of `f(x)=3x^2+2x-5` :
We can put the quadratic in vertex form: `f(x)=a(x-h)^2+k` where the vertex is (h,k).
`f(x)=3x^2+2x-5` First factor out 3 so that the leading coefficient is one:
`f(x)=3(x^2+2/3x-5/3)` Now we take 1/2 of the linear coefficient and square it. This number is then added and subtracted --thus we are adding zero, and not really changing the function:
`f(x)=3(x^2+2/3x+1/9-1/9-5/3)` **`1/2*2/3=1/3;(1/3)^2=1/9` **
Note that `x^2+2/3x+1/9` is a perfect square trinomial --
`f(x)=3(x+1/3)^2-48/9` This is in vertex form so the vertex is at `(-1/3,-48/9)`