Question

1. find the vertex of the parabola.

y = 3x² - 6x - 2
(1,-5)

1. what are the zeros?

Explanation:

Step1: Find x-coordinate of vertex

For $y=ax^2+bx+c$, $x=-\frac{b}{2a}$. Here $a=3$, $b=-6$.
$x=-\frac{-6}{2\times3}=1$

Step2: Find y-coordinate of vertex

Substitute $x=1$ into $y=3x^2-6x-2$.
$y=3(1)^2-6(1)-2=3-6-2=-5$

Step3: Find zeros using quadratic formula

Zeros satisfy $3x^2-6x-2=0$. Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\frac{6\pm\sqrt{(-6)^2-4\times3\times(-2)}}{2\times3}=\frac{6\pm\sqrt{36+24}}{6}=\frac{6\pm\sqrt{60}}{6}=\frac{6\pm2\sqrt{15}}{6}=1\pm\frac{\sqrt{15}}{3}$

Answer:

1. $(1, -5)$
2. $x=1+\frac{\sqrt{15}}{3}$ and $x=1-\frac{\sqrt{15}}{3}$