5. graph the following function by answer the questions and graphing on the grid provided. $ f(x) = \frac{x + 2}{x^2 + 6x - 7} $ $ (x - 1)(x + 1) $ $ \text{a) point of discontinuity(if it exists) or explain why it doesnt} $ $ \text{there are no common factors = no point of disc.} $ $ \text{b) vertical asymptote(s).} $ $ \text{c) horizontal asymptote?} $ $ \text{(will it cross the horizontal asymptote?)} $ $ \text{d) x intercept(s)} $ $ \text{@ 2} $ $ \text{e) y intercept} $ $ \text{f) sign analysis:} $

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Factor denominator
First, factor the denominator $x^{2}+6x - 7$. We need two numbers that multiply to $-7$ and add to $6$. Those numbers are $7$ and $-1$, so $x^{2}+6x - 7=(x + 7)(x - 1)$. The numerator is $x + 2$.
## Step2: Check common factors
Now, check if the numerator ($x + 2$) and the factored denominator ($(x + 7)(x - 1)$) have any common factors. Since $x+2$ is not equal to $x + 7$ or $x - 1$ for all $x$, there are no common factors.
## Step3: Conclusion on discontinuity
A point of discontinuity (hole) occurs when there is a common factor in the numerator and denominator (which gets canceled). Since there are no common factors, there is no point of discontinuity.
# Answer:
There is no point of discontinuity because the numerator $x + 2$ and the factored denominator $(x + 7)(x - 1)$ have no common factors.

### Part (b)
# Explanation:
## Step1: Recall vertical asymptote rule
Vertical asymptotes of a rational function $f(x)=\frac{N(x)}{D(x)}$ occur where $D(x)=0$ and $N(x)
eq0$ (since we already checked for holes).
## Step2: Solve denominator for zero
We have the denominator factored as $(x + 7)(x - 1)$. Set each factor equal to zero:
 - For $x+7 = 0$, we get $x=-7$.
 - For $x - 1=0$, we get $x = 1$.
## Step3: Check numerator at these x - values
Now, check the numerator at $x=-7$ and $x = 1$:
 - When $x=-7$, the numerator $x + 2=-7 + 2=-5
eq0$.
 - When $x = 1$, the numerator $x + 2=1+2 = 3
eq0$.
So, the vertical asymptotes occur at these $x$-values.
# Answer:
The vertical asymptotes are $x=-7$ and $x = 1$.

### Part (c)
# Explanation:
## Step1: Recall horizontal asymptote rules
For a rational function $f(x)=\frac{N(x)}{D(x)}$, where the degree of $N(x)$ is $n$ and the degree of $D(x)$ is $m$:
 - If $n\lt m$, the horizontal asymptote is $y = 0$.
 - If $n=m$, the horizontal asymptote is $y=\frac{\text{leading coefficient of }N(x)}{\text{leading coefficient of }D(x)}$.
 - If $n>m$, there is no horizontal asymptote (but there may be an oblique asymptote).
## Step2: Determine degrees
The degree of the numerator $N(x)=x + 2$ is $n = 1$ (since the highest power of $x$ is $1$). The degree of the denominator $D(x)=x^{2}+6x - 7$ is $m = 2$ (highest power of $x$ is $2$). Since $n=1\lt m = 2$, the horizontal asymptote is $y = 0$.
## Step3: Check if it crosses the asymptote
To check if the graph crosses the horizontal asymptote $y = 0$, we set $f(x)=0$ (since $y = f(x)$) and solve for $x$. So, $\frac{x + 2}{(x + 7)(x - 1)}=0$. A fraction is zero when the numerator is zero (and the denominator is not zero). Set $x+2=0$, we get $x=-2$. Now, check if $x=-2$ makes the denominator zero: $(-2 + 7)(-2 - 1)=(5)(-3)=-15
eq0$. So, when $x=-2$, $y = 0$, which means the graph crosses the horizontal asymptote $y = 0$ at the point $(-2,0)$.
# Answer:
The horizontal asymptote is $y = 0$, and the graph does cross the horizontal asymptote at $x=-2$ (the point $(-2,0)$).

### Part (d)
# Explanation:
## Step1: Recall x - intercept rule
The $x$-intercepts of a function $y = f(x)$ occur where $y = 0$ (i.e., $f(x)=0$) and the function is defined (denominator is not zero).
## Step2: Solve $f(x)=0$
Set $f(x)=\frac{x + 2}{(x + 7)(x - 1)}=0$. A fraction is zero when the numerator is zero (and the denominator is not zero). So, set the numerator equal to zero: $x+2 = 0$, which gives $x=-2$.
## Step3: Check denominator at $x=-2$
Check if $x=-2$ makes the denominator zero: $(-2 + 7)(-2 - 1)=(5)(-3)=-15
eq0$. So, $x=-2$ is a valid $x$-intercept.
# Answer:
The $x$-intercept is at $x=-2$ (the point $(-2,0)$).

### Part (e)
# Explanation:
## Step1: Recall y - intercept rule
The $y$-intercept of a function $y = f(x)$ occurs where $x = 0$ (we substitute $x = 0$ into the function and find $y$).
## Step2: Substitute $x = 0$ into $f(x)$
Substitute $x = 0$ into $f(x)=\frac{x + 2}{x^{2}+6x - 7}$:
$f(0)=\frac{0 + 2}{0^{2}+6(0)-7}=\frac{2}{-7}=-\frac{2}{7}$
# Answer:
The $y$-intercept is at $y=-\frac{2}{7}$ (the point $(0,-\frac{2}{7})$).

### Part (f)
# Explanation:
## Step1: Find critical points
The critical points for sign analysis of a rational function are the $x$-values where the function is zero ( $x=-2$) and where the function is undefined (vertical asymptotes at $x=-7$ and $x = 1$). These points divide the real number line into intervals: $(-\infty,-7)$, $(-7,-2)$, $(-2,1)$, and $(1,\infty)$.
## Step2: Test intervals
 - For the interval $(-\infty,-7)$, let's choose a test point, say $x=-8$.
$f(-8)=\frac{-8 + 2}{(-8 + 7)(-8 - 1)}=\frac{-6}{(-1)(-9)}=\frac{-6}{9}=-\frac{2}{3}\lt0$
 - For the interval $(-7,-2)$, let's choose $x=-3$.
$f(-3)=\frac{-3 + 2}{(-3 + 7)(-3 - 1)}=\frac{-1}{(4)(-4)}=\frac{-1}{-16}=\frac{1}{16}\gt0$
 - For the interval $(-2,1)$, let's choose $x = 0$ (we already calculated $f(0)=-\frac{2}{7}\lt0$)
 - For the interval $(1,\infty)$, let's choose $x = 2$.
$f(2)=\frac{2+2}{(2 + 7)(2 - 1)}=\frac{4}{(9)(1)}=\frac{4}{9}\gt0$
# Answer:
- Interval $(-\infty,-7)$: $f(x)\lt0$
- Interval $(-7,-2)$: $f(x)\gt0$
- Interval $(-2,1)$: $f(x)\lt0$
- Interval $(1,\infty)$: $f(x)\gt0$

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Factor denominator

Step2: Check common factors

Step3: Conclusion on discontinuity

Step1: Recall vertical asymptote rule

Step2: Solve denominator for zero

Step3: Check numerator at these x - values

Step1: Recall horizontal asymptote rules

Step2: Determine degrees

Step3: Check if it crosses the asymptote

Answer:

Part (b)