Question

1. find the vertex of the parabola.

y = 3x² - 6x - 2
(1,-5)

1. what are the zeros?

Explanation:

Step1: Identify coefficients for vertex x

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

Step2: Calculate vertex x-value

$x = \frac{6}{6} = 1$

Step3: Find vertex y-value

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

Step4: Simplify y-value

$y = -5$

Step5: Identify coefficients for zeros

Use quadratic formula: $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$, $a=3$, $b=-6$, $c=-2$

Step6: Compute discriminant

$\Delta = b^2-4ac = (-6)^2 - 4(3)(-2) = 36 + 24 = 60$

Step7: Simplify square root of discriminant

$\sqrt{60} = 2\sqrt{15}$

Step8: Calculate zero values

$x = \frac{6\pm2\sqrt{15}}{6} = \frac{3\pm\sqrt{15}}{3}$

Answer:

Vertex: $(1, -5)$
Zeros: $x=\frac{3+\sqrt{15}}{3}$ and $x=\frac{3-\sqrt{15}}{3}$