Question

question
plot all intercepts and asymptotes of the following rational function. if you get a fraction or decimal then plot as close to the true location as possible.
$f(x) = 3 + \frac{1}{x + 5}$
plot rational function
vertical asymptote horizontal asymptote
x intercept y intercept
click on a button then drag it into place.

Explanation:

Step1: Find Vertical Asymptote

The vertical asymptote of a rational function occurs where the denominator is zero (and the numerator is not zero at that point). For \( f(x)=3 + \frac{1}{x + 5} \), the denominator is \( x + 5 \). Set \( x+5 = 0 \), so \( x=-5 \).

Step2: Find Horizontal Asymptote

For a rational function of the form \( y = \frac{a}{x - h}+k \), the horizontal asymptote is \( y = k \). Here, \( f(x)=3+\frac{1}{x + 5} \), so the horizontal asymptote is \( y = 3 \).

Step3: Find y - intercept

To find the y - intercept, set \( x = 0 \) in the function. \( f(0)=3+\frac{1}{0 + 5}=3+\frac{1}{5}=\frac{15 + 1}{5}=\frac{16}{5}=3.2 \)

Step4: Find x - intercept

To find the x - intercept, set \( y = 0 \) (i.e., \( f(x)=0 \)). So \( 0=3+\frac{1}{x + 5} \). Subtract 3 from both sides: \( - 3=\frac{1}{x + 5} \). Cross - multiply: \( -3(x + 5)=1 \). Expand: \( -3x-15 = 1 \). Add 15 to both sides: \( -3x=16 \). Divide by - 3: \( x=-\frac{16}{3}\approx - 5.333\)

Answer:

• Vertical Asymptote: \( x=-5 \)
• Horizontal Asymptote: \( y = 3 \)
• y - intercept: \( y=\frac{16}{5}=3.2 \) (or the point \( (0,3.2) \))
• x - intercept: \( x =-\frac{16}{3}\approx - 5.333 \) (or the point \( (-\frac{16}{3},0) \))