## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$b$
1. If the major axis of the parabola is parallel to the $y$-axis then it has the form of $(x=h)^2=4a(y-k)$. where $(h, k)$ is the vertex of the parabola and $4a$ is a constant which will be positive if the graph opens up and negative if it opens down. 2. If the major axis of the parabola is parallel to the x-axis, then it has the form of $4a(x-h)=(y-k)^2$, where $(h, k)$ is the vertex of the parabola and $4a$ is a constant which will be positive if the graph opens right and negative if it opens left. Since the major axis is parallel to the $y$-axis and opens up, in all cases $4a=\pm4$ and the vertex is: $(0,0)$. Thus, the equation becomes : $4(y-0)=(x-0)^2 \implies x^2=4y$ So, our answer is $\bf{b}$.