find the vertex of the parabola f(X)=3x^2+5x-2

Question

you have to find the vertex.

Answer 1

Given parabola f(x) = y = 3x^2+5x-2

Compare it standard parabola form y = ax^2+bx+c.

a = 3, b = 5, c = -2.

Axis of symmetry x = -b/2a  = -5/6

To find vertex substitute x value in given equation.

y = 3(-5/6)^2+5(-5/6)-2

y = 3(25/36)-25/6-2

y = 25/12-25/6-2

y = (25-50-24)/12

y = -49/12

Vertex of parabola = (-5/6,-49/12) = (-0.83,-4.08)