Question

1. if $f(x)=x^2 + 4x + 4$, then the domain of $y = \frac{1}{f(x)}$ is

a. $\\{x|x \
eq -2, 2, x \in \mathrm{r}\\}$
b. $\\{x|x \
eq -2, x \in \mathrm{r}\\}$
c. $\\{x|x \
eq 2, x \in \mathrm{r}\\}$
d. $\\{x|x \in \mathrm{r}\\}$

1. the graph of $y = \frac{1}{14x + 90}$ has $a$ horizontal asymptotes, $b$ vertical asymptotes, and the graphs of the functions $y = 14x + 90$ and $y = \frac{1}{14x + 90}$ will intersect $c$ times. the values of $a$, $b$, and $c$ are ______.

1. sketch the graph of $y = \frac{1}{3x - 6}$, outlining the steps used.

Explanation:

Question 3

Step1: Factor the denominator

$f(x)=x^2+4x+4=(x+2)^2$

Step2: Find undefined points

Set $(x+2)^2=0$, solve: $x=-2$

Step3: Define the domain

All real numbers except $x=-2$

Step1: Find horizontal asymptote (A)

For $y=\frac{1}{14x+90}$, as $x\to\pm\infty$, $y\to0$. So $A=1$.

Step2: Find vertical asymptote (B)

Set $14x+90=0$, solve: $x=-\frac{90}{14}=-\frac{45}{7}$. So $B=1$.

Step3: Find intersection count (C)

Set $14x+90=\frac{1}{14x+90}$. Let $t=14x+90$, so $t^2=1$, $t=\pm1$.

• $14x+90=1 \implies 14x=-89 \implies x=-\frac{89}{14}$
• $14x+90=-1 \implies 14x=-91 \implies x=-\frac{13}{2}$

Two valid solutions, so $C=2$.

Step1: Identify undefined points

Set $3x-6=0$, solve: $x=2$. Vertical asymptote: $x=2$.

Step2: Find horizontal asymptote

As $x\to\pm\infty$, $y=\frac{1}{3x-6}\to0$. Horizontal asymptote: $y=0$.

Step3: Plot key points

• For $x>2$: $x=3$, $y=\frac{1}{3(3)-6}=\frac{1}{3}$; $x=4$, $y=\frac{1}{6}$
• For $x<2$: $x=1$, $y=\frac{1}{3(1)-6}=-\frac{1}{3}$; $x=0$, $y=-\frac{1}{6}$

Step4: Sketch the graph

Draw asymptotes $x=2$ (dashed vertical) and $y=0$ (dashed horizontal). Plot the points, then draw two curves: one in the right half ($x>2$, above $y=0$) and one in the left half ($x<2$, below $y=0$), approaching the asymptotes.

Answer:

B. $\{x|x
eq -2, x \in \mathbb{R}\}$

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Question 4