Rational Functions: Graphs and Critical Features (613)

Identify Vertical and Horizontal Asymptotes based on the graph

• V.A. --> x = 0

  H.A. --> y = 1

• V.A. --> x = 1

  H.A. --> y = 0

• V.A. --> y = 1

  H.A. --> x = 0

• V.A. --> x = 0

  H.A. --> none

Identify Vertical and Horizontal Asymptotes based on the graph

• V.A. --> x = -1

  H.A. --> y = -1

• V.A. --> x = -1

  H.A. --> y = -2

• V.A. --> y = -1

  H.A. --> x = 2

• V.A. --> x = -1

  H.A. --> y = 2

Identify Vertical and Horizontal Asymptotes based on Graph...

• V.A. --> x = 1/2

  H.A. --> y = -1

• V.A. --> x = -1/2

  H.A. --> y = 1

• V.A. --> y = -1

  H.A. --> x = 1/2

• V.A. --> x = 1

  H.A. --> y = -1

Identify any removable discontinuities (holes) in this rational function:

 $f\left(x\right)=\ \frac{\left(x-2\right)\left(x+3\right)}{\left(x+2\right)}$  

• No holes

• Hole at (2, 0) and (-3, 0)

• Hole at x = - 2

• Hole at (-2,0)

Identity any removable discontinuities (holes) in this rational function...

 $g\left(x\right)\ =\ \frac{9x^2-1}{3x+1}$  

• hole at (-1/3, 1/3)

• no hole

• hole at (-1/3, 0)

• hole at (-1/3, -2)

What are the x - intercepts of this rational function?

 $h\left(x\right)=\frac{\left(x-2\right)\left(x+1\right)}{x^2}$  

• There are none

• There are two: (2, 0) and (-1, 0)

• There are three, x = 0, x = 2, x = -1

• There are two: (0, 2) and (0, -1)

Tags

CCSS.HSA.APR.B.3

 $f\left(x\right)\ =\ \frac{2x^2-4}{x+2}$  

What is the y - intercept of this rational function?

• (0, 2)

• (-2, 0)

• There are two:  y = +/- 2

• (0, -2)

Tags

CCSS.HSF.IF.B.4

CCSS.HSF.IF.C.7

 $g\left(x\right)=\frac{x^3-1}{\left(x^2+x+1\right)}$  

What are the x - intercepts of this rational function?

• (0, 1)

• (1, 0)

• -1, and 1

• 1, - 1, and 0

What is the y - intercept of this rational function?

 $h\left(x\right)\ =\ \frac{3x^2-7x+4}{\left(x-1\right)^2}$  

• (4/3, 0) ... (1, 0) is not in domain of function

• (4, 0)

• (0, 4)

• (0, -4)

Identify any Vertical Asymptotes for this rational function...

 $f\left(x\right)\ =\ \frac{\left(x+2\right)\left(x+2\right)}{\left(x+2\right)}$  

• x  =  - 2

• y  =  - 2

• there are none

• (- 2, 0)