Question

y = 3x^2
plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph - a - function button.

Explanation:

Step1: Find the vertex

For a parabola of the form $y = ax^{2}+bx + c$, in $y = 3x^{2}$, $a = 3$, $b = 0$, $c = 0$. The $x$ - coordinate of the vertex is $x=-\frac{b}{2a}=0$. Substituting $x = 0$ into $y = 3x^{2}$, we get $y = 0$. So the vertex is $(0,0)$.

Step2: Find left - hand points

Let $x=-1$, then $y = 3\times(-1)^{2}=3$. Let $x=-2$, then $y = 3\times(-2)^{2}=12$. So the left - hand points are $(-1,3)$ and $(-2,12)$.

Step3: Find right - hand points

Let $x = 1$, then $y = 3\times1^{2}=3$. Let $x = 2$, then $y = 3\times2^{2}=12$. So the right - hand points are $(1,3)$ and $(2,12)$.

Answer:

Vertex: $(0,0)$; Left - hand points: $(-1,3),(-2,12)$; Right - hand points: $(1,3),(2,12)$