How can polygons be considered a subcategory of two-dimensional figures?

Mathematics

1 answer:

Greeley [361]4 years ago

6 0

Well, the polygon has at least three straight sides and angles, typically 5 or more

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Part 4: Identify the vertex, focus and directrix of each. Then sketch the graph.

Nezavi [6.7K]

Answer: 1) Vertex: (6, -2) Focus: (6, -7/4) Directrix: y = -9/4

2) Vertex: (-2, -1) Focus: (-7/4, -1) Directrix: x = -9/4

<u>Step-by-step explanation:</u>

Rewrite the equation in vertex format y = a(x - h)² + k or x = a(y - k)² + h by completing the square. Divide the b-value by 2 and square it - add that value to both sides of the equation.

(h, k) is the vertex
p is the distance from the vertex to the focus
-p is the distance from the vertex to the directrix

$\bullet\quad a=\dfrac{1}{4p}$

1) y = x² - 12x + 34

$y-34=x^2-12x\\\\y-34+\bigg(\dfrac{-12}{2}\bigg)^2=x^2-12x+\bigg(\dfrac{-12}{2}\bigg)^2\\\\\\y-34+36=(x-6)^2\\\\y+2=(x-6)^2\\\\y=(x-6)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(6,-2)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(6,-\dfrac{7}{4}\bigg)\\\\\\\text{Directrix: y= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad y=-\dfrac{9}{4}$

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2) x = y² + 2y - 1

$x+1=y^2+2y\\\\x+1+\bigg(\dfrac{2}{2}\bigg)^2=y^2+2y+\bigg(\dfrac{2}{2}\bigg)^2\\\\\\x+1+1=(y+1)^2\\\\x+2=(y+1)^2\\\\x=(y+1)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(-2,-1)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(-\dfrac{7}{4},-1\bigg)\\\\\\\text{Directrix: x= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=-\dfrac{9}{4}$