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Worksheet: Interpreting the Relative Maximum and Minimum and the End Behavior of a Graph

Q1:

Find the maximum or minimum value of the function , given .

  • A The minimum value is 0.
  • B The maximum value is 1.
  • C The maximum value is 0.
  • D The minimum value is 1.
  • E The minimum value is 3.

Q2:

Consider the graph of the quartic shown.

Which of the points 𝐴 , 𝐡 , and 𝐢 is a local maximum?

  • A 𝐢
  • B 𝐴
  • C 𝐡

Which of the points 𝐴 , 𝐡 , and 𝐢 is a local minimum?

  • A 𝐴 and 𝐢
  • B 𝐡 and 𝐢
  • C 𝐴 and 𝐡

Which of the points 𝐴 , 𝐡 , and 𝐢 is a global minimum?

  • A 𝐡
  • B 𝐴
  • C 𝐢

By considering the end behavior of the quartic, determine whether the leading coefficient is positive or negative.

  • ANegative
  • BPositive

Q3:

Consider the cubic graph shown.

What are the coordinates of the local maximum?

  • A ( 0 , βˆ’ 1 )
  • B ( 3 , βˆ’ 2 )
  • C ( βˆ’ 1 , 0 )
  • D ( βˆ’ 2 , 3 )

What are the coordinates of the local minimum?

  • A ( βˆ’ 2 , 3 )
  • B ( 3 , βˆ’ 2 )
  • C ( βˆ’ 1 , 0 )
  • D ( 0 , βˆ’ 1 )

If we consider the end behavior of this cubic, it enters into the bottom left quadrant and exits from the top right quadrant. Is the leading coefficient of this cubic positive or negative?

  • ANegative
  • BPositive

Q4:

What is the maximum value of the function 𝑦 = 5 βˆ’ | 2 βˆ’ π‘₯ | ?

Q5:

Consider a function 𝑓 ( π‘₯ ) = π‘Ž π‘₯ + 𝑏 𝑛 , where π‘Ž , 𝑏 , and 𝑛 are integers larger than 1. Which of the following statements is true?

  • A There will be neither an absolute maximum nor an absolute minimum.
  • B If 𝑛 is odd, there will be an absolute maximum.
  • C The existence of extremes cannot be determined without more information.
  • DIf 𝑛 is even, there will be an absolute minimum.

Q6:

Which of the following has the lowest minimum value?

  • A π‘˜ ( π‘₯ ) = 1 3 ( π‘₯ βˆ’ 3 ) ( π‘₯ + 4 )
  • B 𝑓 ( π‘₯ ) = π‘₯ βˆ’ 4 π‘₯ 2
  • Ca quadratic function β„Ž whose graph cuts the π‘₯ -axis at βˆ’1 and 2 and cuts the 𝑦 -axis at βˆ’4
  • D
    π‘₯ 0 1 2 3 4 5 6 7
    𝑔 ( π‘₯ ) 6 0 βˆ’4 βˆ’6 βˆ’6 βˆ’4 0 6
  • E 𝑓 ( π‘₯ ) = 5 ( π‘₯ βˆ’ 3 ) + 4 2