Question

graph the parabola.
y = \frac{1}{4}x^{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.

Explanation:

Step1: Find the vertex

For the parabola $y = ax^{2}+bx + c$, in $y=\frac{1}{4}x^{2}$, $a=\frac{1}{4},b = 0,c = 0$. The $x$ - coordinate of the vertex is $x=-\frac{b}{2a}=0$. Substitute $x = 0$ into $y=\frac{1}{4}x^{2}$, we get $y = 0$. So the vertex is $(0,0)$.

Step2: Find left - hand points

Let $x=-2$, then $y=\frac{1}{4}\times(-2)^{2}=1$. Let $x=-4$, then $y=\frac{1}{4}\times(-4)^{2}=4$. So two left - hand points are $(-2,1)$ and $(-4,4)$.

Step3: Find right - hand points

Let $x = 2$, then $y=\frac{1}{4}\times2^{2}=1$. Let $x = 4$, then $y=\frac{1}{4}\times4^{2}=4$. So two right - hand points are $(2,1)$ and $(4,4)$.

Answer:

Vertex: $(0,0)$; Left - hand points: $(- 4,4),(-2,1)$; Right - hand points: $(2,1),(4,4)$