Question

Part 3. Identify the vertex, focus, and directrix of each. Then sketch the graph.

Answer 1

Answer:  $\bold{1.\quad \text{Vertex}=(-1,5)\qquad \text{Focus}=\bigg(-\dfrac{3}{2},5\bigg)\qquad \text{Directrix}:x=-\dfrac{1}{2}}$

               $\bold{2.\quad \text{Vertex}=(-4,-1)\quad \text{Focus}=\bigg(-4,-\dfrac{1}{4}\bigg)\qquad \text{Directrix}:y=-\dfrac{7}{4}}$

Step-by-step explanation:

The vertex form of a parabola is x = a(y - k)² + h  or  y = a(x - h)² + k

• (h, k) is the vertex

    $\bullet\quad a=\dfrac{1}{4p}$

• p is the distance from the vertex to the focus
• -p is the distance from the vertex to the directrix

1)

$x=-\dfrac{1}{2}(y-5)^2-1\qquad \rightarrow \qquad a=-\dfrac{1}{2}\quad (h,k)=(-1,5)\\\\a=\dfrac{1}{4p}\qquad \rightarrow \quad -\dfrac{1}{2}=\dfrac{1}{4p}\qquad \rightarrow \quad-2=4p\qquad \rightarrow \quad-\dfrac{1}{2}=p\\\\\\\text{Focus = Vertex + p}\\\\.\qquad =\dfrac{-2}{2}+\dfrac{-1}{2}\\\\.\qquad =-\dfrac{3}{2}\qquad \rightarrow \qquad \text{Focus}=\bigg(-\dfrac{3}{2},5\bigg)$

$\text{Directrix: x = Vertex - p}\\\\.\qquad \qquad x=\dfrac{-2}{2}-\dfrac{-1}{2}\\\\.\qquad \qquad x=-\dfrac{1}{2}$

***********************************************************************************

2)

$y=\dfrac{1}{3}(x+4)^2-1\qquad \rightarrow \qquad a=\dfrac{1}{}\quad (h,k)=(-4,-1)\\\\a=\dfrac{1}{4p}\qquad \rightarrow \quad \dfrac{1}{3}=\dfrac{1}{4p}\qquad \rightarrow \quad3=4p\qquad \rightarrow \quad\dfrac{3}{4}=p\\\\\\\text{Focus = Vertex + p}\\\\.\qquad =\dfrac{-4}{4}+\dfrac{3}{4}\\\\.\qquad =-\dfrac{1}{4}\qquad \rightarrow \qquad \text{Focus}=\bigg(-4,-\dfrac{1}{4}\bigg)$

$\text{Directrix: y = Vertex - p}\\\\.\qquad \qquad y=\dfrac{-4}{4}-\dfrac{3}{4}\\\\.\qquad \qquad y=-\dfrac{7}{4}$