Answered: Trace the curve y = (x-1)(x+3)"

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Step 1

Given curve is

$y = \frac{x}{(x - 1) (x + 3)}$

1.Domain of the given curve is $(- \infty, - 3) \cup (- 3, 1) \cap (1, \infty)$

2.Extend: $x < - 3 o r - 3 < x < 1 o r x > 1$

3.Intercepts

3(a).X intercept

Put y =0

$0 = \frac{x}{(x - 1) (x + 3)} \Rightarrow x = 0$

3(b).Y intercept

Put x = 0

$y = \frac{0}{(0 - 1) (0 + 3)} \Rightarrow y = 0$

4. Origin

Given curve passes through the origin (0,0)

Step 2

5. Symmetry

The given curve has no symmetry

6. Asymptotes

6(a).Vertical asymptote

Find the root of the denominator which is not the root of the numerator

$x - 1 = 0 \Rightarrow x = 1 x + 3 = 0 \Rightarrow x = - 3$

Therefore x = 1 and x = -3 are the vertical asymptotes

6(b).Horizontal asymptote

$\lim_{x \to \infty} y = \lim_{x \to \infty} \frac{x}{(x - 1) (x + 3)} = \lim_{x \to \infty} \frac{x}{x^{2} (1 - \frac{1}{x}) (1 + \frac{3}{x})} = \lim_{x \to \infty} \frac{1}{x (1 - \frac{1}{x}) (1 + \frac{3}{x})} = \frac{1}{\infty (1 - \frac{1}{\infty}) (1 + \frac{3}{\infty})} = 0$

Therefore, y=0 is the horizontal asymptote of the given curve

6(c).Slant asymptote

There is no slant in this curve, since the degree of the denominator is greater than the degree of the numerator

Step 3

7. Monotonicity

To check the monotonicity find the derivative of the given curve

$y = \frac{x}{(x - 1) (x + 3)}$

$y' = \frac{d}{d x} (\frac{x}{(x - 1) (x + 3)}) = \frac{\frac{d}{d x} [x] . (x - 1) (x + 3) - x \frac{d}{d x} [(x - 1) (x + 3)]}{{((x - 1) (x + 3))}^{2}} = \frac{(x - 1) (x + 3) - x [(x - 1) + (x + 3)]}{{((x - 1) (x + 3))}^{2}} = \frac{(x - 1) (x + 3) - x [2 x + 2]}{{((x - 1) (x + 3))}^{2}} = \frac{x^{2} - x + 3 x - 3 - 2 x^{2} - 2 x}{{((x - 1) (x + 3))}^{2}} y' = \frac{- x^{2} + 3}{{((x - 1) (x + 3))}^{2}}$

Critical points are x = 1 and x = -3

Monotone interval behavior

Intervals	$(- \infty, - 3)$	$(- 3, 1)$	$(1, \infty)$
Test value	-4	0	2
Sign of y'	y'<0	y'<0	y'<0
Conclusion	Decreasing	Decreasing	Decreasing

The given graph is monotonically decreasing. And doesn't have intercepts other than origin

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