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makvit [3.9K]
3 years ago
15

Write functions for each of the following transformations using function notation. Choose a different letter to represent each f

unction. For example, you can use R to represent rotations. Assume that a positive rotation occurs in the counterclockwise direction.
translation of a units to the right and b units up
reflection across the y-axis
reflection across the x-axis
rotation of 90 degrees counterclockwise about the origin, point O
rotation of 180 degrees counterclockwise about the origin, point O
rotation of 270 degrees counterclockwise about the origin, point O

please use this format:
T(x, y) = (x + 6, y)
T<6, 0>(x, y) = (x + 6, y)
T6, 0(x, y) = (x + 6, y)
Mathematics
1 answer:
kupik [55]3 years ago
6 0

Answer:

(a)\ T_{a,b} =(x +a,y+b)

(b)\ F_{y\ axis} = (-x,y)

(c)\ F_{x\ axis} = (x,-y)

(d)\ R_{o,90^o} = (-y,x)

(e)\ R_{o,180^o} = (-x,-y)

(f)\ R_{o,270^o} = (y,-x)

Step-by-step explanation:

Solving (a): Translate a units right, b units up

When a function is translated a units right, the number of units will be added to the x coordinate

When a function is translated b units up, the number of units will be added to the y coordinate.

So, we have:

T_{a,b} =(x +a,y+b)

Solving (b): Reflect across y-axis

When a function is translated across the y-axis, the x coordinate gets negated.

So, we have:

F_{y\ axis} = (-x,y)

Solving (c): Reflect across x-axis

When a function is translated across the x-axis, the y coordinate gets negated.

So, we have:

F_{x\ axis} = (x,-y)

Solving (d): 90 degrees rotation counterclockwise

When a function is rotated 90 degrees counterclockwise, the y-coordinates gets negated and then swapped with the x-coordinate.

So, we have:

R_{o,90^o} = (-y,x)

Solving (e): 180 degrees rotation counterclockwise

When a function is rotated 180 degrees counterclockwise, the coordinates are negated.

So, we have:

R_{o,180^o} = (-x,-y)

Solving (f): 270 degrees rotation counterclockwise

When a function is rotated 270 degrees counterclockwise, the x-coordinates gets negated and then swapped with the y-coordinate.

So, we have:

R_{o,270^o} = (y,-x)

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Solving quadratic equations, it is found that he needs to charge:

1. He needs to charge $40 to break even.

2. He needs to charge $30 for a profit of $600.

<h3>What is a quadratic function?</h3>

A quadratic function is given according to the following rule:

y = ax^2 + bx + c

The solutions are:

  • x_1 = \frac{-b + \sqrt{\Delta}}{2a}
  • x_2 = \frac{-b - \sqrt{\Delta}}{2a}

In which:

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The profit equation in this problem is:

P(x) = -3x² + 150x - 1200.

He breaks even when P(x) = 0, hence:

-3x² + 150x - 1200 = 0.

The coefficients are a = -3, b = 150, c = -1200, hence:

  • \Delta = 150^2 - 4(-3)(-1200) = 8100
  • x_1 = \frac{-150 + \sqrt{8100}}{-6}
  • x_2 = \frac{-150 - \sqrt{8100}}{-6} = 40

He needs to charge $40 to break even.

For a profit of $600, we have that P(x) = 600, hence:

-3x² + 150x - 1200 = 600.

-3x² + 150x - 1800 = 0.

The coefficients are a = -3, b = 150, c = -1800, hence:

  • \Delta = 150^2 - 4(-3)(-1800) = 900
  • x_1 = \frac{-150 + \sqrt{900}}{-6}
  • x_2 = \frac{-150 - \sqrt{900}}{-6} = 30

He needs to charge $30 for a profit of $600.

More can be learned about quadratic equations at brainly.com/question/24737967

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