Question

Use the information to write the vertex form and standard form of each parabola for vertex: (-3,-6), focus (- 119/40, -6)

Answer 1

Step 1:

If you have the equation of a parabola in vertex form

$y=a(x-h)^2\text{ + k}$

then the vertex is at (h,k) and the focus is (h,k+14a). Notice that here we are working with a parabola with a vertical axis of symmetry, so the x-coordinate of the focus is the same as the x-coordinate of the vertex.

Step 2:

$\begin{gathered} (y-h)^2\text{ = }4p(x\text{ - k)} \\ S\text{ince the vertex is (-3 , -6) and the focus is below the vertex} \\ p\text{ = -6 - (-6) = 0} \\ \\ h\text{ = -6 and k = -}3 \\ (y+6)^2\text{ = 4}\times\frac{-239}{40}\times(x\text{ +3)} \\ 10(y+6)^2\text{ = }-239(x+3) \\ \end{gathered}$

Step 3

$\begin{gathered} \text{Vertex form} \\ 10(y+6)^2\text{ = -239(x + 3)} \end{gathered}$

Step 4:

In standard form

$\begin{gathered} \frac{-10}{239}(y+6)^2\text{ - 3 = x} \\ \text{x = }\frac{-10}{239}(y+6)^2\text{ - 3} \end{gathered}$