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geniusboy [140]
2 years ago
9

Can someone please take the time out of they’re day and help me and with this question

Mathematics
1 answer:
Bumek [7]2 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

(a)

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (0, 100) and (x₂, y₂ ) = (2, 200) ← 2 points on the line

m = \frac{200-100}{2-0} = \frac{100}{2} = 50

The line crosses the y- axis at (0, 100 ) ⇒ c = 100

y = 50x + 100 ← equation of line

where y is monthly savings and x the time in months

(b)

Slope is the monthly deposit , that is $50 per month

The y- intercept indicates the original amount in her account , that is $100

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Read 2 more answers
(01.03. 106, 1.08 HC)
zlopas [31]

Answer:

  A) see attached for a graph. Range: (-∞, 7]

  B) asymptotes: x = 1, y = -2, y = -1

  C) (x → -∞, y → -2), (x → ∞, y → -1)

Step-by-step explanation:

<h3>Part A</h3>

A graphing calculator is useful for graphing the function. We note that the part for x > 1 can be simplified:

  \dfrac{-x^2+x+2}{x^2-3x+2}=-\dfrac{(x-2)(x+1)}{(x-2)(x-1)}=-\dfrac{x+1}{x-1}\quad x\ne 2

This has a vertical asymptote at x=1, and a hole at x=2.

The function for x ≤ 1 is an ordinary exponential function, shifted left 1 unit and down 2 units. Its maximum value of 3^-2 = 7 is found at x=1.

The graph is attached.

The range of the function is (-∞, 7].

__

<h3>Part B</h3>

As we mentioned in Part A, there is a vertical asymptote at x = 1. This is where the denominator (x-1) is zero.

The exponential function has a horizontal asymptote of y = -2; the rational function has a horizontal asymptote of y = (-x/x) = -1. The horizontal asymptote of the exponential would ordinarily be y=0, but this function has been translated down 2 units.

__

<h3>Part C</h3>

The end behavior is defined by the horizontal asymptotes:

  for x → -∞, y → -2

  for x → ∞, y → -1

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