Answer:
[tex]y=\frac{1}{12}(x+2)^{2}+1[/tex]
Explanation:
Given a parabola with the following properties:
• Vertex: (-2, 1)
,
• Focus: (-2, 4)
We want to write an equation for the parabola.
The standard equation of an up-facing parabola with a vertex at (h,k) and a focal length |p| is given as:
[tex]\begin{equation}(x-h)^{2}=4 p(y-k)\end{equation}[/tex][tex]\begin{gathered} Vertex,(h,k)=(-2,1)\implies h=-2,k=1 \\ Focus,(h,k+p)=(-2,4)\implies h=-2,k+p=4 \end{gathered}[/tex]
We solve for p:
[tex]\begin{gathered} k+p=4 \\ 1+p=4 \\ p=4-1 \\ p=3 \end{gathered}[/tex]
Substitute the values h=-2, k=1, and p=3 into the standard form given earlier:
[tex]\begin{gathered} (x-(-2))^2=4(3)(y-1) \\ (x+2)^2=12(y-1) \\ \text{ Divide both sides by 12} \\ \frac{1}{12}(x+2)^2=y-1 \\ \text{ Add 1 to both sides of the equation} \\ y=\frac{1}{12}(x+2)^2+1 \end{gathered}[/tex]
The equation for the parabola is:
[tex]y=\frac{1}{12}(x+2)^{2}+1[/tex]
The last option is correct.
