Question

What is the approximate value of the function at x = 1? graph inside --->

Answer 1

@jim_thompson5910

Answer 2

look at the graph

locate x = 1 on the x axis and see where it lands on the curve

Answer 3

the value of the function is equal to the y coordinate of the point at x = 1

Answer 4

that sentence just killed my brain cells

Answer 5

what's not making sense?

Answer 6

wait wait wait. so the curve lands on -0.75 correct? because once it touches that box that's where its closest to

Answer 7

yeah roughly, you are correct

Answer 8

when x = 1, y is y = -0.75 roughly

Answer 9

do you know how to find the range and domain of functions

Answer 10

yes I do

Answer 11

find the range of
\[f (x) = \frac{ 2 }{ x - 3 } - 5\]

Answer 12

how would you find the horizontal asymptote of that

Answer 13

i have no idea... don't think i've learned that

Answer 14

oh wait, I didn't simplify first, hold on

Answer 15

\[f (x) = \frac{ 2 }{ x - 3 } - 5\]


\[f (x) = \frac{ 2 }{ x - 3 } - 5\frac{x-3}{x-3}\]


\[f (x) = \frac{ 2 }{ x - 3 } - \frac{5(x-3)}{x-3}\]


\[f (x) = \frac{ 2 }{ x - 3 } - \frac{5x-15}{x-3}\]


\[f (x) = \frac{ 2 - (5x-15) }{ x - 3 }\]


\[f (x) = \frac{ 2 - 5x+15 }{ x - 3 }\]


\[f (x) = \frac{ -5x + 17 }{ x - 3 }\]


Since the degree of the numerator equals the degree of the denominator (they are both equal to 1), this means that the horizontal asymptote is y = -5/1 = -5

Answer 16

the range in interval notation is (-infinity, -5) U (-5, infinity)

Answer 17

i think i understand. now what about the domain of a different problem

Answer 18

\[f (x) = \frac{ 1 }{ x + 1 } - 4\]

Answer 19

hint: you cannot divide by zero

Answer 20

now that i know how to find the range i need to know how to find the domain.

Answer 21

if you cannot divide by zero, what does that mean for the domain?

Answer 22

no solution?

Answer 23

let's say the denominator was zero, so that would mean that x+1 = 0

what is x?

Answer 24

-1

Answer 25

therefore, if x = -1, then x+1 is zero

Answer 26

which means that you can plug in any value you want for x...BUT...x cannot be -1

Answer 27

x cannot be -1 to avoid division by zero

everything else is just fine

Answer 28

so the domain is all real numbers except -1

Answer 29

exactly

Answer 30

thank you! alright next question,

Answer 31

\[\frac{ 2x - 5 }{ x - 2 } - 2 = \frac{ 3 }{ x + 2 }\]

Answer 32

\[\frac{ 2x - 5 }{ x - 2 } - 2 = \frac{ 3 }{ x + 2 }\]


\[\frac{ 2x - 5 }{ x - 2 } - 2\frac{x-2}{x-2} = \frac{ 3 }{ x + 2 }\]


\[\frac{ 2x - 5 }{ x - 2 } - \frac{2(x-2)}{x-2} = \frac{ 3 }{ x + 2 }\]


\[\frac{ 2x - 5 }{ x - 2 } - \frac{2x-4}{x-2} = \frac{ 3 }{ x + 2 }\]


\[\frac{ 2x - 5-(2x-4)}{x-2} = \frac{ 3 }{ x + 2 }\]


\[\frac{ 2x - 5-2x+4}{x-2} = \frac{ 3 }{ x + 2 }\]


I'll let you finish

Answer 33

@jim_thompson5910 i got x = 1

Answer 34

you are correct

Answer 35

thank god you literally saved my life

Answer 36

not really lol, but I'm glad to be of help

Answer 37

i do have another if you wanna stick around
\[\frac{ x - 4 }{ 4 } + \frac{ x }{ 3 } = 6\]

Answer 38

@jim_thompson5910

Answer 39

I'll start you off


\[\frac{ x - 4 }{ 4 } + \frac{ x }{ 3 } = 6\]


\[\frac{ 3*(x - 4) }{ 3*4 } + \frac{ 4x }{ 4*3 } = 6\]


\[\frac{ 3x - 12 }{ 12 } + \frac{ 4x }{ 12 } = 6\]


\[\frac{ 3x - 12 + 4x }{ 12 } = 6\]

Answer 40

can you cancel -12 and 12 to get rid of your denominator?

Answer 41

no

Answer 42

that is not a valid algebraic move

Answer 43

x = 12

Answer 44

good

Answer 45

now i have another domain question

Answer 46

\[f(x) = \frac{ 2 }{ x + 1 } - 3\]

Answer 47

tell me what you get

Answer 48

all real numbers except -1

Answer 49

you are correct

Answer 50

then what is the domain of \[f (x) = - \frac{ 3 }{ x - 4 } + 2\]

Answer 51

would it be all real numbers except 4?

Answer 52

very good

Answer 53

you got the hang of this domain stuff

Answer 54

yessss okay next one

Answer 55

\[\frac{ 3 }{ n - 1 } + \frac{ 2 }{ 3 } = \frac{ 5 }{ 3 }\]

Answer 56

\[\large \frac{ 3 }{ n - 1 } + \frac{ 2 }{ 3 } = \frac{ 5 }{ 3 }\]


\[\large \frac{ 3*3 }{ 3(n - 1) } + \frac{ 2(n-1) }{ 3(n-1) } = \frac{ 5 }{ 3 }\]


\[\large \frac{ 9 }{ 3(n - 1) } + \frac{ 2n-2 }{ 3(n-1) } = \frac{ 5 }{ 3 }\]


\[\large \frac{ 9 + 2n-2 }{ 3(n-1) } = \frac{ 5 }{ 3 }\]

Answer 57

n = 4

Answer 58

correct

Answer 59

you are the bomb

Answer 60

i have 3 questions left

Answer 61

\[\frac{ 4 }{ n } - \frac{ 1 }{ 2 } = -\frac{ 6 }{ 3n }\]

Answer 62

\[\large \frac{ 4 }{ n } - \frac{ 1 }{ 2 } = -\frac{ 6 }{ 3n }\]


\[\large \frac{ 2*4 }{ 2*n } - \frac{ n*1 }{ n*2 } = -\frac{ 6 }{ 3n }\]


\[\large \frac{ 8 }{ 2n } - \frac{ n }{ 2n } = -\frac{ 6 }{ 3n }\]


\[\large \frac{ 8 - n }{ 2n } = -\frac{ 6 }{ 3n }\]

Answer 63

i got 12n = 24n - 3n..... they all have n's

Answer 64

\[\large \frac{ 4 }{ n } - \frac{ 1 }{ 2 } = -\frac{ 6 }{ 3n }\]


\[\large \frac{ 2*4 }{ 2*n } - \frac{ n*1 }{ n*2 } = -\frac{ 6 }{ 3n }\]


\[\large \frac{ 8 }{ 2n } - \frac{ n }{ 2n } = -\frac{ 6 }{ 3n }\]


\[\large \frac{ 8 - n }{ 2n } = -\frac{ 6 }{ 3n }\]


\[\large 3n(8 - n) = -6*2n\]


\[\large 24n - 3n^2 = -12n\]


does that help?

Answer 65

now what the heck am i supposed to do

Answer 66

\[\large 24n - 3n^2 = -12n\]

\[\large 24n - 3n^2 + 12n = 0\]

\[\large -3n^2 + 36n = 0\]

\[\large -3n(n - 12) = 0\]

how about now?

Answer 67

don't forget about checking your answers and keep the domain in mind

Answer 68

n = 0 and n = 12?

Answer 69

check both of those

only one is the true answer

Answer 70

12

Answer 71

good

Answer 72

i have 2 left and they're super quick graph ones

Answer 73

What is the approximate value of the function at x = 0?

Answer 74

you tell me

Answer 75

i don't know...4?

Answer 76

you're looking at the point when y = 0

Answer 77

how about x = 0

Answer 78

1.3 ?

Answer 79

close enough

Answer 80

last one! What is the approximate value of the function at x =5

Answer 81

again tell me what you got

Answer 82

this one i really don't know.

Answer 83

find x = 5 on the x axis

Answer 84

draw a vertical line through x = 5

find where that vertical line crosses the red curve

Answer 85

-0.75?

Answer 86

then report the y coordinate

Answer 87

yes approximately

Answer 88

i got every single one right... i actually wanna cry. thank you beyond words. @jim_thompson5910

Answer 89

you're welcome