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Lera25 [3.4K]
2 years ago
6

Find the slope and the y-intercept of the linear function shown below

Mathematics
1 answer:
Leviafan [203]2 years ago
7 0

Answer:

You give no Problem or answers to choose from

Step-by-step explanation:

Sorry

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Answer:  \bold{1.\quad \text{Vertex}}:(-3,-6)\qquad \text{Focus}:\bigg(-3,-\dfrac{17}{4}\bigg)\qquad \text{Directrix}:y=-\dfrac{19}{4}

               2.  Vertex: (-5, 4)           Focus: (-6, 4)                 Directrix: x = -4

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

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

  • (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 = (x + 3)² - 6        →       a = 1      (h, k) = (-3, -6)

a=\dfrac{1}{4p}\qquad \rightarrow \qquad 1=\dfrac{1}{4p} \qquad \rightarrow \qquad p=\dfrac{1}{4}\\\\\text{Focus = Vertex + p}\\\\.\qquad =\dfrac{-18}{4}+\dfrac{1}{4}\\\\.\qquad = -\dfrac{17}{4}\qquad \rightarrow \qquad \text{Focus}=\bigg(-3,-\dfrac{17}{4}\bigg)\\\\\\\text{Directrix: y = Vertex - p}\\\\.\qquad \quad y=\dfrac{-18}{4}-\dfrac{1}{4}\\\\.\qquad \quad y= -\dfrac{19}{4}

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

2.\quad x=-\dfrac{1}{4}(y-4)^2-5\qquad \rightarrow \quad a=\dfrac{1}{4}\qquad (h, k)=(-5,4)

a=\dfrac{1}{4p}\qquad \rightarrow \qquad -\dfrac{1}{4}=\dfrac{1}{4p} \qquad \rightarrow \qquad p=-1\\\\\text{Focus = Vertex + p}\\\\.\qquad =-5+\ -1\\\\.\qquad = -6\qquad \rightarrow \qquad \text{Focus}=(-6,4)\\\\\\\text{Directrix: y = Vertex - p}\\\\.\qquad \quad y=-5+\ -1\\\\.\qquad \quad y= -4

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