Qacstion $\newline$ Determine each feature of the graph of the given function. $\newline$ $f(x)=\frac{5x+5}{2x+2}$ $\newline$ Answer sthempti cut of $2$ $\newline$ Horizontal Asymptote: $\newline$ $y=$ $\newline$ Vertical Asymptote: $\newline$ $x=$ $\newline$ \left[\begin{array}{cc}$\newline\text{x-Intercept: } & (\square,0)\square (\newline$\text{y-Intercept: } & (0,\square)\text{ Noytares } $\newline$ \end{array}\right]\) $\newline$ Hole: $\newline$ $\square,\square$ Ssb bolk

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Algebra 2

Domain and range of quadratic functions: equations

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Q. Qacstion

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Determine each feature of the graph of the given function.

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f(x)=\frac{5x+5}{2x+2}

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Answer sthempti cut of

2

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Horizontal Asymptote:

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y=

\newline

Vertical Asymptote:

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x=

\newline

\left[\begin{array}{cc}$\newline\text{x-Intercept: } & (\square,0)\square (\newline$\text{y-Intercept: } & (0,\square)\text{ Noytares }

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\end{array}\right]\)

\newline

Hole:

\newline

\square,\square

Ssb bolk

Simplify the Function: First, simplify the function $f(x)$ to see if it can be reduced. $f(x) = \frac{5x+5}{2x+2}$ can be simplified by factoring out a common factor of $5$ from the numerator and $2$ from the denominator. $f(x) = \frac{5(x+1)}{2(x+1)}$ We can cancel out the $(x+1)$ terms since they are common to the numerator and the denominator. $f(x) = \frac{5}{2}$
Horizontal Asymptote: Now that we have simplified the function, we can determine the horizontal asymptote. Since the function simplifies to a constant, $f(x) = \frac{5}{2}$ , the horizontal asymptote is $y = \frac{5}{2}$ .
Vertical Asymptote: Next, we look for the vertical asymptote. $\newline$ Originally, the function had a denominator of $2x+2$ . Setting the denominator equal to zero gives us the vertical asymptote. $\newline$ $2x + 2 = 0$ $\newline$ $2x = -2$ $\newline$ $x = -1$ $\newline$ So, the vertical asymptote is $x = -1$ .
X-Intercept: To find the x-intercept, we set $f(x)$ to zero and solve for $x$ .
$0 = \frac{5x+5}{2x+2}$
This equation suggests that the numerator must be zero for the function to be zero.
$5x + 5 = 0$
$5x = -5$
$x = -1$
However, $x = -1$ is also the vertical asymptote, which means there is no x-intercept because the function is undefined at $x = -1$ .
Y-Intercept: To find the y-intercept, we set $x$ to zero and solve for $f(x)$ . $f(0) = \frac{5(0)+5}{2(0)+2}$ $f(0) = \frac{5}{2}$ So, the y-intercept is $(0, \frac{5}{2})$ .
Check for Holes: Finally, we check for any holes in the graph. $\newline$ A hole occurs where the original function is undefined due to a common factor in the numerator and denominator that was canceled out. $\newline$ Since we canceled $(x+1)$ , there is a hole at $x = -1$ . $\newline$ To find the $y$ -coordinate of the hole, we substitute $x = -1$ into the reduced function $f(x) = \frac{5}{2}$ . $\newline$ The $y$ -coordinate of the hole is also $\frac{5}{2}$ . $\newline$ So, the hole is at $(-1, \frac{5}{2})$ .