\(y = 3x^2+6x-5\)
For a parabola \(ax^2+bx+c\) the vertex's x equals:
\(\frac{-b}{a}\)
\(a = 3, b =6\)
\(x = \frac{-6}{2*3}\)
refine:
\(x = -1\)
Plug in x = -1 to find the y value :
\(y=3(-1)^2+6(-1)-5\)
refine :
\(y = -8\)
Therefore the parabola vertex is :
\((-1,-8)\)
if a < 0 , then the vertex is a maximum value
if a > 0, then then vertex is a minimum value
a=3
Therefore \((-1,-8)\) minimum
hope this helps!
