Question

What is the vertex of the graph of the equation y = 3x²+6x+1?

Answer 1

y=3x²+6x+1
to find the x value of the vertex, find the axis of symmetry which is -b/2a where a=3 and b=6,
so the axis of symmetry
=-6/2(3)
=-1
to find the y value of the vertex sub the axis of symmetry which is -1 back into the original equation.
y= 3(-1)²+6(-1)+1
y=-2

therefore the vertex is (-1,-2)

Answer 2

`y=3x^2+6x+1\\\\V(-(b)/(2a),-(\Delta)/(4a))\\\\-(b)/(2a)=-(6)/(2\cdot3)=-(6)/(6)=-1\\\\\Delta=6^2-4\cdot3\cdot1=36-12=24\\\\-(\Delta)/(4a)=-(24)/(4\cdot3)=-(24)/(12)=-2\\\\V(-1,-2)`