Question

1. $y = x^2$ vertex: ____ $a = $ ____

Explanation:

Step1: Identify vertex form

The standard vertex form of a parabola is $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex. For $y=x^2$, this is $y=1(x-0)^2+0$.

Step2: Extract vertex and $a$

From the vertex form, $h=0$, $k=0$, so vertex is $(0,0)$. The coefficient $a=1$.

Step3: Plot key points

Calculate points for the parabola:

• When $x=-2$, $y=(-2)^2=4$
• When $x=-1$, $y=(-1)^2=1$
• When $x=0$, $y=0^2=0$
• When $x=1$, $y=1^2=1$
• When $x=2$, $y=2^2=4$

Plot these points and draw a smooth curve through them.

Answer:

Vertex: $(0, 0)$
$a = 1$
(The parabola is plotted by connecting the points $(-2,4)$, $(-1,1)$, $(0,0)$, $(1,1)$, $(2,4)$ with a smooth U-shaped curve.)