Question

Which is the vertex of the graph of y=3x^2+2x+1

Answer 1

Answer:

$(-\frac{1}{3},2)$

Step-by-step explanation:

The x-coordinate of the vertex is $x=-\frac{b}{2a}$

Therefore, $x=-\frac{2}{2(3)}=-\frac{2}{6}=-\frac{1}{3}$.

Now, the y-coordinate can be found by plugging in the x-coordinate into the function:

$y=3x^2+2x+1$

$y=3(-\frac{1}{3})^2+2(\frac{1}{3})+1$

$y=3(\frac{1}{9})+\frac{2}{3}+1$

$y=\frac{3}{9}+\frac{2}{3}+1$

$y=\frac{1}{3}+\frac{2}{3}+1$

$y=\frac{3}{3}+1$

$y=1+1$

$y=2$

So, the vertex of the graph is $(-\frac{1}{3},2)$