the graph shows a parabola and its focus. write the equation of the parabola in vertex form.

Answer:

$y = \frac{1}{8}(x - 0)^2+0$

Explanation:

Step1: Identify the vertex

The vertex of the parabola is at the origin $(0,0)$.

Step2: Recall the vertex - form of a parabola

The vertex - form of a parabola opening vertically is $y=a(x - h)^2 + k$, where $(h,k)$ is the vertex. Here $h = 0,k = 0$, so the equation is $y=ax^2$.

Step3: Find the value of $a$

The distance $p$ from the vertex $(0,0)$ to the focus $(- 4,0)$ is $|p|$. For a parabola of the form $y = ax^2$, the relationship between $a$ and $p$ is $a=\frac{1}{4p}$. The parabola opens to the left, and the distance from the vertex $(0,0)$ to the focus $(-4,0)$ is $p=- 4$. Then $a=\frac{1}{4\times(-4)}=-\frac{1}{16}$ for a parabola opening left. But if we consider the standard vertical - opening form and use the fact that for a parabola $y = ax^2$ and the properties of focus and directrix, we can also note that for a parabola with vertex at the origin and passing through points symmetrically, we can use another approach. If we assume the general vertical - opening form $y=a(x - 0)^2+0$. We can take a point on the parabola, say $(2, \frac{1}{2})$. Substituting $x = 2$ and $y=\frac{1}{2}$ into $y=ax^2$, we get $\frac{1}{2}=a\times2^2$. Solving for $a$: [ \begin{align*} \frac{1}{2}&=4a\ a&=\frac{1}{8} \end{align*} ] So the equation of the parabola in vertex - form is $y=\frac{1}{8}(x - 0)^2+0$.