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

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

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)

the vertex is (-1,2)

oh, i thought you didn’t put a negative in front of 2. sorry lol

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