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

RELATED: