Math / Chem

Math

Chem

Chem 7.31 - 7.45

HW 3.5 Q4

As $x \rightarrow -3^-, f(x) \rightarrow$ __

The notation $x \rightarrow -3^-$ symbolizes “ $x$ approaches $-3$ from the left”.

Observing the graph, as $x$ approaches $-3$ from the left ( $x\rightarrow-3^-$ ), the function values increase without bound. That is, the function values approach infinity.

Therefore, as $x \rightarrow -3^-, f(x) \rightarrow \infty$

HW 3.5 Q5

$x \rightarrow 3^-, f(x) \rightarrow \infty$

HW 3.5 Q6

$x \rightarrow -\infty, f(x) \rightarrow 0$

HW 3.5 Q9

$h(x) = \frac{x}{x(x+9)}$

Vertical asymptote is at $x=-9$

Hole at $x=0$

HW 3.5 Q10

$f(x) = \frac{x-6}{x^2-13x+42}$

$= \frac{x-6}{(x-6)(x-7)}$

Vertical asymptotes: $x=7$

$(x-6)$ cancels out.

$x-6=0$
$x = 6$

Hole at $x=6$

HW 3.5 Q11

$f(x) = \frac{18x}{7x^2+2}$

The degree of the numerator $(n)$ is $1$ and the degree of the denominator $(m)$ is $2$ .

If $n<m$ , the $x$ -axis ( $y=0$ ) is the horizontal asymptote.

Horizontal asymtote is $\color{blue}{y=0}$ .

HW 3.5 Q12

$f(x) = \frac{-8x+5}{7x+2}$

$n=m$

$y=\frac{a_n}{b_m}$

$y=\frac{-8}{7}$

HW 3.5 Q13

$f(x) = \frac{x^2-4}{x}$

Find the slant asymptote.

$\frac{x^2-4}{x} = x-\frac{4}{x}$

Slant asymtpote is $y=x$ .

$f(-x) = \frac{(-x)^2-4}{(-x)}$

$-f(x) = -\frac{x^2-4}{x}$

$f(-x) = -f(x)$

The graph of $f(x)$ has origin symmetry.

$f(0) = \frac{(0)^2-4}{(0)}$

Since the demoninator is $0$ and dividing anything by $0$ is undefined, we know the graph has no $y$ -intercept.

To find the $x$ -intercept, set the numerator to $0$ .

$x^2-4=0$

Factor:

$(x-2)(x+2)$

$x=-2,x=2$

The $x$ -intercept are $x=-2$ and $x=2$ .

Find the vertical asymptote by setting the demoninator to $0$ :

$x = 0$

Since $n>m$ , there is no horizontal asymptote.

HW 3.5 Q14

$f(x) = \frac{x^2+6}{x+1}$

Algebraic Long division:

          x  - 1
         ------------
(x + 1) | x² + 0x + 6
         -x² - 1x
         ------------
              -1x + 6
              +1x + 1
              -------
                    7

$y=x-1$

$f(0) = \frac{(0)^2+6}{(0)+1}$

$= 6/1 = 6$

$y$ -intercept occurs at $(0,6)$ .

$x^2+6=0$
$x^2=-6$
$x = \sqrt{-6}$

Since $x$ is an imaginary number, there is no $x$ -intercept.

HW 3.5 Q15

$f(x) = \frac{x^2-x-2}{x-4}$

Synthetic division:

4 | 1 - 1 - 2
    1   4  12 
   ------------
    1   3  10

Slant asymptote: $x+3$

$f(-x) = \frac{x^2+x-2}{-x-4}$

$= \frac{(x-1)(x+2)}{-x-4}$

$-f(x) = -\frac{x^2-x-2}{x-4}$

$f(-x) \ne f(x)$
$f(-x) \ne -f(x)$

$y$ -intercept $= f(0)$

$f(0) = \frac{0-0-2}{0-4}$

$=-2/-4$

$=1/2$

$x$ -intercept: $x^2-x-2=0$

Factor:

$(x+1)(x-2)=0$

$x$ -intercepts: $\color{blue}{x=-1,x=2}$

Vertical asymptote: $x-4=0$

$x=4$