## anonymous one year ago Piecewise function question

1. misty1212

$f(x) = |x+4| = \left\{\begin{array}{rcc} x + 4 & \text{if} & x \geq -4 \\ - x - 4& \text{if} & x < -4 \end{array} \right.$

2. anonymous

yes i have a graph that i really need help on

3. anonymous

could you please hhelp me out

4. misty1212

you need to graph something?

5. anonymous

no ive got a graph and have some questions about it

6. misty1212

ok i can try to help, post the picture

7. anonymous

i can post a picture of the worksheet so it'll be easier to see the graph

8. anonymous

9. anonymous

10. anonymous

this is the graph and the first question is f(1)=

11. anonymous

and i got 3 as the answer

12. misty1212

oh i thought you had to find the function

13. anonymous

i do later on

14. misty1212

yeah $$f(1)=3$$

15. anonymous

and the next one is f(-11)

16. anonymous

but on the graph its only till -10

17. misty1212

then we better find the function for $$x\leq -5$$

18. misty1212

you got the slope right, it is $$\frac{1}{2}$$

19. anonymous

yes that is the slope

20. misty1212

goes through the point $$(-5,3)$$ so it is $y+5=\frac{1}{2}(x-3)$

21. misty1212

you can plug 11 in to that if you like

22. anonymous

oh alright that makes sense

23. misty1212

or else write it as $y=\frac{1}{2}(x-3)+5$

24. anonymous

is it -2

25. misty1212

whoah did i screw that up!!

26. anonymous

im not sure haha

27. misty1212

slope is $$\frac{1}{2}$$ point is $$(-5,3)$$ equation is $y-3=\frac{1}{2}(x+5)$ doe

28. anonymous

so its y= 1/2(x+5)+3?

29. misty1212

or $y=\frac{x}{2}+\frac{11}{2}$

30. misty1212

whatever you like

31. anonymous

alright let me solve one sec

32. misty1212

cool henna btw

33. anonymous

thank you :)

34. misty1212

did you get zero yet?

35. anonymous

yepp

36. anonymous

okay so the next one says find x when f(x)=0

37. anonymous

is it 5/2

38. misty1212

lol you just did it

39. anonymous

im not supposed to use decimals

40. misty1212

you found $$f(-11)=0$$ right?

41. anonymous

yes i did

42. misty1212

so that makes "find x when f(x)=0" rather obvious !!

43. anonymous

lol idk why i put 5/2

44. misty1212

plug in $$-11$$ get out zero !!

45. misty1212

me neither lol

46. anonymous

i got it!

47. misty1212

ok next...

48. anonymous

alright then lol on to the next one

49. anonymous

i have to find the x intercepts

50. misty1212

ok we got one right? one is $$-11$$

51. anonymous

and i got (5/2,0) and (-11,0)

52. misty1212

there is another as well, because it is coming back up at the end and that little arrow indicates it continues up

53. misty1212

lets just find the damn function and be done with it k?

54. anonymous

lol alright

55. anonymous

it actually says to write in point slope form and restricted domain in inequality form

56. anonymous

so basically its asking me to write a fuction for each piece

57. misty1212

right so far we are at $f(x) = \left\{\begin{array}{rcc} y-3=\frac{1}{2}x+5 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1 \end{array} \right.$

58. misty1212

what is the slope of the next piece? it is a little hard for me to see it

59. misty1212

maybe $$-2$$?

60. anonymous

you said in the first function theres a 3 ?

61. anonymous

under where you write the 5 in the first equation

62. misty1212

it is a constant (horizontal line) between -5 and 1 right?

63. anonymous

yes thats correct

64. misty1212

ooh damn i made a typo there forgot the parentheses

65. misty1212

$f(x) = \left\{\begin{array}{rcc} y-3=\frac{1}{2}(x+5) & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1 \end{array} \right.$

66. anonymous

im a little confused

67. misty1212

ok lets go slow and make sure you get it

68. anonymous

under the first equation you put a 3

69. anonymous

but im not sure why

70. misty1212

yeah we are not done you see how you have 4 long lines in your answer space?

71. misty1212

because this piecewise function has four distinct parts

72. anonymous

yes i do

73. misty1212

4 different lines

74. anonymous

okay yes i understand now thank you :)

75. misty1212

on the interval from $$-5$$ to $$1$$ the function is a constant, it is $$f(x)=3$$ a horizontal line

76. anonymous

yes

77. misty1212

so the second line down on the left you should have $$3,-5\leq x\leq 1$$ because on that interval it is 3

78. anonymous

i understand

79. anonymous

oh okay

80. misty1212

the next line has slope $$-2$$ and goes through $$(1,3)$$ so its equation is $y=y-3=-2(x-1)$ in point slope form

81. anonymous

would the slope not be a 0?

82. misty1212

yes, the slope is zero, the equation is $$y=3$$

83. anonymous

okay i got it

84. anonymous

okay i got it

85. misty1212

$f(x) = \left\{\begin{array}{rcc} \frac{1}{2}(x+5)+3 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1\\ -2(x-1)+3 &\text{if}&1<x<6 \end{array} \right.$

86. misty1212

one more to go

87. anonymous

okay i got it

88. misty1212

last one also has slope $$\frac{1}{2}$$ right?

89. anonymous

yes it does

90. misty1212

and goes through the point $$(6,-7$$?

91. anonymous

yes i think so

92. misty1212

$f(x) = \left\{\begin{array}{rcc} \frac{1}{2}(x+5)+3 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1\\ -2(x-1)+3 &\text{if}&1<x<6\\ \frac{1}{2}(x-6)-7&\text{if}&x\geq6 \end{array} \right.$

93. misty1212

that should do it

94. anonymous

yes i got it i just have a couple of small questions that i wanted to confirm

95. misty1212

ok

96. anonymous

are the x intercepts (5/2,0) and (-11,0)?

97. misty1212

yes, but there is one more

98. misty1212

set $\frac{1}{2}(x-7)-7=0$ for to find the last one

99. misty1212

aggh!!

100. misty1212

$\frac{1}{2}(x-6)-7=0$

101. misty1212

that is the part on the right think you get $$20$$ if i did it right

102. anonymous

i was solving it one sec ill confirm

103. anonymous

yes i got 20

104. anonymous

so itll be (20,0)

105. misty1212

yeah

106. anonymous

okay and the range is (-infinity, infintiy)?

107. misty1212

yeah that is what the arrows indicate

108. anonymous

and the positives are (-11, 5/2)?

109. anonymous

thast the last one haha

110. misty1212

i don't know what that means

111. misty1212

oh you mean the interval over which it is positive ?

112. anonymous

the positives as in the parts in the upper part of the graph

113. anonymous

yes

114. misty1212

$(-11,\frac{5}{2})$ is one interval but there is another

115. misty1212

since it is zero at 20, and it keeps going up, the other one is $$(20,\infty)$$

116. anonymous

alrighty thank you so much

117. anonymous

im just putting the functions in slope intercept form now and the doamins in interval notation

118. anonymous

ill be done in a sec if you have time to check those

119. misty1212

ok good sure

120. anonymous

y= 1/2x+11/2

121. anonymous

y=3 y=-2x+5 y=1/2x-10

122. misty1212

slope intercept for the first one yes

123. anonymous

yes i am doing slope intercept for all of them

124. misty1212

ok let me scroll way up and see i know the first two are right

125. misty1212

third is right too

126. anonymous

okay annddd the last one

127. misty1212

yeah that also looks good

128. anonymous

okay and the interval notation is (infinity, -5)? for the first one?

129. anonymous

[-5,-1] (1,6) (infinity, 6]

130. misty1212

ooh mistake there

131. misty1212

second interval should be $$[-5,1]$$

132. misty1212

and also do not put infinity on the left on the last interval!!

133. anonymous

i think thats what i said for the second one

134. misty1212

should be on the right, go from least to greatest i.e. $$(6,\infty)$$

135. anonymous

is the first interbal correct?

136. misty1212

$$(-\infty,-5)$$ yes

137. anonymous

and for the last one do i use a bracket before 6 since its greater than or equal to

138. misty1212

since it meets up there it is your choice

139. misty1212

btw usually these piecewise functions are written with inequalities not intervals but if you are supposed to use intervals that is fine too

140. anonymous

yeah it asked us to write it nboth wwyas

141. anonymous

do you think i should use a bracket?

142. misty1212

kk makes no difference

143. anonymous

alrigh thank you :)

144. misty1212

yw ms kumar lol $\huge \color\magenta\heartsuit$