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Helen [10]
3 years ago
10

Select "Growth" or "Decay" to classify each function.

Mathematics
1 answer:
allochka39001 [22]3 years ago
5 0

1. The growth rate equation has a general form of:

y = A (r)^t

The function is growth when r≥1, and it is a decay when r<1. Therefore:

y=200(0.5)^2t                    --> Decay

y=1/2(2.5)^t/6                  --> Growth

y=(0.65)^t/4                       --> Decay

 

2. We rewrite the given equation (1/3)^d−5 = 81

 

Take the log of both sides:

(d – 5) log(1/3) = log 81

d – 5 = log 81 / log(1/3)

d – 5 = - 4

 

Multiply both sides by negative 1:

- d + 5 = 4

So the answer is D

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

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  • 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}

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

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}

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