Question

1. What sequence of transformations will yield the graph of g(x) from the graph of f(x)? (7 pts.) f(x) = x g(x) = -3(x + 4)3 2. Determine the vertical asymptote(s), horizontal asymptote, and oblique asymptote of the function. If none, write "NONE.” (6 pts.) x2 +3x+2 f(x) = x-1 1). Ver. Asy. 2). Hor. Asy. 3) Obl. Asy.

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Answer 1

Solution:

(1)

Start with f(x)=x³

1) Horizontal shift of |h| units
• f(x) → f(x + h)
h=4 unit moves left

f(x + 4) = (1 + 4) f(3) = (x + 4)

2) Vertical Stretch by a factor of a
• f(x) → af(x)

a=3 then

$\\f(x)=3(x+4)^3$

3) Reflection across x-axis
• f(x) → -f(x)

$\\f(x)=-3(x+4)^3$

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(2)

$f(x)=\frac{x^{2} + 3 x + 2}{x - 1}$

VERTICAL ASYMPTOTES

The line x=Lx=L is a vertical asymptote of the function

$f(x)=\frac{x^{2} + 3 x + 2}{x - 1}$

if the limit of the function (one-sided) at this point is infinite.

In other words, it means that possible points are points where the denominator equals 0 or doesn't exist.

So, find the points where the denominator equals 0 and check them.

x=1, check:

lim 1+1+ + 3.0 + 2 - 1 = 0

Since the limit is infinite, then x=1 is a vertical asymptote

HORIZONTAL ASYMPTOTES

Line y=Ly=L is a horizontal asymptote of the function y=f(x)y=f(x), if either

$\lim_{x \to \infty} f{\left(x \right)}=L$

or,

$\lim_{x \to- \infty} f{\left(x \right)}=L$

and L is finite.

Calculate the limits:

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

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

Thus, there are no horizontal asymptotes.

SLANT ASYMPTOTES

Do polynomial long division

$\frac{x^{2} + 3 x + 2}{x - 1}=x + 4 + \frac{6}{x - 1}$

The rational term approaches 0 as the variable approaches infinity.

Thus, the slant asymptote is

y=x+4

Vertical asymptote: x=1

No horizontal asymptotes.

Slant asymptote: y=x+4