Question

$y = 3x^2 - 5$
$x = \frac{-b}{2a}$
axis of symmetry:
vertex:

Explanation:

Step1: Identify quadratic coefficients

The quadratic function is $y=3x^2 - 5$, which matches the standard form $y=ax^2+bx+c$. Here, $a=3$, $b=0$, $c=-5$.

Step2: Calculate axis of symmetry

Use formula $x=\frac{-b}{2a}$.
$\frac{-0}{2\times3}=0$

Step3: Find vertex y-coordinate

Substitute $x=0$ into $y=3x^2-5$.
$y=3(0)^2 - 5=-5$

Step4: Generate x-y table

Choose $x=-2,-1,0,1,2$ and compute $y$:

• $x=-2$: $y=3(-2)^2-5=12-5=7$
• $x=-1$: $y=3(-1)^2-5=3-5=-2$
• $x=0$: $y=-5$
• $x=1$: $y=3(1)^2-5=3-5=-2$
• $x=2$: $y=3(2)^2-5=12-5=7$

Answer:

Axis of Symmetry: $x=0$
Vertex: $(0, -5)$
x-y Table:

$x$	$y$
-1	-2
0	-5
1	-2
2	7

(Note: To graph, plot these points and draw a symmetric parabola opening upwards with vertex at $(0,-5)$)