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blondinia [14]
3 years ago
11

D(x)=(x^2-12x+20)/(3x)

Mathematics
1 answer:
krok68 [10]3 years ago
6 0

Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

<u>Explanation:</u>

d(x) = \frac{x^{2}-12x+20}{3x}

      = \frac{(x-2)(x - 10)}{3x}

Discontinuities: (terms that cancel out from numerator and denominator):

Nothing cancels so there are NO discontinuities.

Vertical asymptote (denominator cannot equal zero):

3x ≠ 0  

<u>÷3</u>   <u>÷3 </u>

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

degree of numerator (2) > degree of denominator (1)

so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

  •        <u>(1/3)x - 4    </u>
  •    3x)    x² - 12x + 20
  •             <u>x²        </u>
  •                  -12x
  •                  <u>-12x         </u>
  •                             20 (stop! because there is no "x")

So, slant asymptote is to be drawn at (1/3)x - 4



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What is the equation for a line that passes through the points (5,-4) and (-10,17
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Answer:

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Step-by-step explanation:

Here the given points are ( 5, -4) & ( -10, 17) -

Equation of a line whose points are given such that

( x_{1},y_{1} ) & ( x_{2}, y_{2}  )-

 y - y_{1}   = \frac{y_{2} - y_{1}  }{x_{2} - x_{1}}  ( x - x_{1}  )

i.e.  <em>y - (-4)= \frac{17 - (-4)}{-10 - 5}  ( x- 5)</em>

<em>      y + 4 = \frac{17 + 4}{ -15} ( x - 5)</em>

<em>      y + 4 = \frac{21}{-15}  ( x - 5 )</em>

<em>      ( y + 4)  =  \frac{7}{- 5} ( x - 5)</em>

<em>      5 (y + 4 ) = - 7 (x - 5 )</em>

<em>      5 y + 20 = -7 x + 35</em>

<em>      7 x + 5 y = 15</em>

Hence the equation of the required line whose passes trough the points ( 5, -4) & ( -10, 17)  is 7 x + 5 y = 15.

7 0
3 years ago
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