Question

secondary math iii // module 4
rational expressions & functions - 4.3

1. $y = \frac{x^2}{(x + 6)(x - 6)}$

graph of the function
degree of num. ____ degree of denom. ____
equation of horizontal asymptote:
equation of vertical asymptote(s):
y - intercept: (write as a point)
x - intercept(s): (write as points)

1. $y = \frac{(x - 6)}{x + 3}$

graph of the function
degree of num. ____ degree of denom. ____
equation of horizontal asymptote:
equation of vertical asymptote(s):
y - intercept: (write as a point)
x - intercept(s): (write as points)

1. $y = \frac{10x}{(x + 3)^2}$

graph of the function
degree of num. ____ degree of denom. ____
equation of horizontal asymptote:
equation of vertical asymptote(s):
y - intercept: (write as a point)
x - intercept(s): (write as points)

1. $y = \frac{(x + 1)}{(x + 2)(x - 5)}$

graph of the function
degree of num. ____ degree of denom. ____
equation of horizontal asymptote:
equation of vertical asymptote(s):
y - intercept: (write as a point)
x - intercept(s): (write as points)

Explanation:

Problem 10: \( y = \frac{x^2}{(x + 6)(x - 6)} \)

Step 1: Degree of Numerator and Denominator

• The numerator is \( x^2 \), so its degree is 2.
• The denominator is \( (x + 6)(x - 6) = x^2 - 36 \), so its degree is 2.

Step 2: Horizontal Asymptote

For a rational function \( \frac{f(x)}{g(x)} \), if the degrees of \( f(x) \) and \( g(x) \) are equal, the horizontal asymptote is \( y = \frac{\text{leading coefficient of } f(x)}{\text{leading coefficient of } g(x)} \). Here, leading coefficients are both 1, so horizontal asymptote is \( y = 1 \).

Step 3: Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is not zero). Set \( (x + 6)(x - 6) = 0 \), so \( x = -6 \) and \( x = 6 \).

Step 4: y - intercept

To find the y - intercept, set \( x = 0 \). Then \( y=\frac{0^2}{(0 + 6)(0 - 6)}=\frac{0}{- 36}=0 \). So the y - intercept is \( (0,0) \).

Step 5: x - intercepts

To find x - intercepts, set \( y = 0 \). So \( \frac{x^2}{(x + 6)(x - 6)}=0 \), which implies \( x^2 = 0 \), so \( x = 0 \). The x - intercept is \( (0,0) \).

Problem 11: \( y=\frac{x - 6}{x + 3} \)

Step 1: Degree of Numerator and Denominator

• The numerator is \( x - 6 \), degree is 1.
• The denominator is \( x + 3 \), degree is 1.

Step 2: Horizontal Asymptote

Since degrees are equal, horizontal asymptote is \( y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}=\frac{1}{1}=1 \).

Step 3: Vertical Asymptote

Set denominator \( x + 3 = 0 \), so \( x=-3 \).

Step 4: y - intercept

Set \( x = 0 \), \( y=\frac{0 - 6}{0+3}=\frac{-6}{3}=-2 \). So y - intercept is \( (0,-2) \).

Step 5: x - intercepts

Set \( y = 0 \), \( \frac{x - 6}{x + 3}=0 \), so \( x - 6 = 0 \), \( x = 6 \). x - intercept is \( (6,0) \).

Problem 12: \( y=\frac{10x}{(x + 3)^2} \)

Step 1: Degree of Numerator and Denominator

• Numerator: \( 10x \), degree is 1.
• Denominator: \( (x + 3)^2=x^2 + 6x+9 \), degree is 2.

Step 2: Horizontal Asymptote

Since degree of numerator (1) < degree of denominator (2), the horizontal asymptote is \( y = 0 \).

Step 3: Vertical Asymptote

Set \( (x + 3)^2=0 \), so \( x=-3 \).

Step 4: y - intercept

Set \( x = 0 \), \( y=\frac{10(0)}{(0 + 3)^2}=0 \). y - intercept is \( (0,0) \).

Step 5: x - intercepts

Set \( y = 0 \), \( \frac{10x}{(x + 3)^2}=0 \), so \( 10x = 0 \), \( x = 0 \). x - intercept is \( (0,0) \).

Problem 13: \( y=\frac{x + 1}{(x + 2)(x - 5)} \)

Step 1: Degree of Numerator and Denominator

• Numerator: \( x + 1 \), degree is 1.
• Denominator: \( (x + 2)(x - 5)=x^2-3x - 10 \), degree is 2.

Step 2: Horizontal Asymptote

Since degree of numerator (1) < degree of denominator (2), horizontal asymptote is \( y = 0 \).

Step 3: Vertical Asymptotes

Set \( (x + 2)(x - 5)=0 \), so \( x=-2 \) and \( x = 5 \).

Step 4: y - intercept

Set \( x = 0 \), \( y=\frac{0 + 1}{(0 + 2)(0 - 5)}=\frac{1}{-10}=-\frac{1}{10} \). y - intercept is \( (0,-\frac{1}{10}) \).

Step 5: x - intercepts

Set \( y = 0 \), \( \frac{x + 1}{(x + 2)(x - 5)}=0 \), so \( x + 1 = 0 \), \( x=-1 \). x - intercept is \( (-1,0) \).

Answer:

s

Problem 10

• Degree of num.: 2; Degree of denom.: 2
• Equation of horizontal asymptote: \( y = 1 \)
• Equation of vertical asymptote(s): \( x=-6, x = 6 \)
• y - intercept: \( (0,0) \)
• x - intercept(s): \( (0,0) \)

Problem 11

• Degree of num.: 1; Degree of denom.: 1
• Equation of horizontal asymptote: \( y = 1 \)
• Equation of vertical asymptote(s): \( x=-3 \)
• y - intercept: \( (0,-2) \)
• x - intercept(s): \( (6,0) \)

Problem 12

• Degree of num.: 1; Degree of denom.: 2
• Equation of horizontal asymptote: \( y = 0 \)
• Equation of vertical asymptote(s): \( x=-3 \)
• y - intercept: \( (0,0) \)
• x - intercept(s): \( (0,0) \)

Problem 13

• Degree of num.: 1; Degree of denom.: 2
• Equation of horizontal asymptote: \( y = 0 \)
• Equation of vertical asymptote(s): \( x=-2, x = 5 \)
• y - intercept: \( (0,-\frac{1}{10}) \)
• x - intercept(s): \( (-1,0) \)