Question

PLEASE HELP!!!! FAST I need the vertical + horizontal asymptotes for f(x)=2x^2+7x-15/5x+x^2

Answer 1

HA y=2 and va x-0 and x=5

Answer 2

H.A.: y = 2, correct

V.A. x = 0, correct

V.A. x = -5 ... notice how it's negative

Answer 3

x^2 + 5x = 0

x(x+5) = 0

x = 0 or x+5 = 0

x = 0 or x = -5

so that explains why it's x = -5 and not x = 5

Answer 4

thank you so much!!! I wrote the wrong this on here haha

Answer 5

wrote what wrong

Answer 6

I need to check my work on my last problem. Would you be able to check it? I did it a million times today and took me hours/

Answer 7

sure go for it

Answer 8

a-f

Answer 9

Thank you so much... Literally working on it for hours

Answer 10

ok post what you did

Answer 11

I got a cp of 6.1 about and concavities of all positive and increasing after 6.1 and decreasing before 6.1.

Answer 12

aan IP at 4

Answer 13

second derivative is 12x^2(x-4)

Answer 14

I couldn't synthetically divide it

Answer 15

stressing me out

Answer 16

ok one sec

Answer 17

kk

Answer 18

f(x) = x^4-8x^3-16x+5

f ' (x) = 4x^3-24x^2-16

f '' (x) = 12x^2-48x

Answer 19

The critical point is roughly 6.10724315175795

Answer 20

you get this from solving 4x^3-24x^2-16 = 0 for x

Answer 21

f(x) is increasing on the interval (6.10724315175795, infinity)

this is a rough approximation of course

Answer 22

f(x) is decreasing on the interval (-infinity, 6.10724315175795)

and both of these facts can be seen from a graph or by using the first derivative test

Answer 23

wait..when I drew the graph it looked concave up

Answer 24

it is concave up

Answer 25

well for the most part it is

Answer 26

oh lol the negative infinity one confused me

Answer 27

actually it's concave down on the interval (0,4)

Answer 28

it passes through 0,$?

Answer 29

12x^2-48x = 0

12x(x-4) = 0

12x = 0 or x-4 = 0

x = 0 or x = 4

Answer 30

0,4

Answer 31

the possible inflection points are at x = 0 and x = 4

Answer 32

but only 4 works right

Answer 33

where is it concave up and down between what intervals

Answer 34

Now see if there is a sign change as you move through x = 0

f '' (x) = 12x^2-48x

f '' (-1) = 12(-1)^2-48(-1)

f '' (-1) = 60


f '' (x) = 12x^2-48x

f '' (1) = 12(1)^2-48(1)

f '' (1) = -36

this means that x = 0 is definitely an inflection point

Answer 35

what are the concavity intervals

Answer 36

notice how f '' (-1) = 60 is positive, so it's concave up on (-infinity, 0)

Answer 37

f '' (1) = -36 is negative, so it's concave down on (0,4)

Answer 38

now plug in a value to the right of x = 4

f '' (x) = 12x^2-48x

f '' (5) = 12(5)^2-48(5)

f '' (5) = 60

f '' is once again positive, so this confirms the other point of inflection is at x = 4 and it's concave up on (4, infinity)

Answer 39

the confusing part was the 0,4

Answer 40

that's the interval where f is concave down

Answer 41

is the graph rising from 6.1,infinity

Answer 42

yeah it's increasing on this interval

Answer 43

and decreasing from -infinity,6.1

Answer 44

yep

Answer 45

would you be able to draw the graph for me lol. I have ridiculous numbers

Answer 46

I plugged 6.1 in and got like -315

Answer 47

I need the graph and the range and I'll be ok

Answer 48

posted the graph

Answer 49

that's the graph of f(x)

Answer 50

the range will be the set of y values that are larger than the smallest possible y value (ie the absolute min)

Answer 51

can you help me lable this.. my teacher wants certain points on here and I didn't understand how he got them

Answer 52

I am taking a month long course it is so accelerated for me

Answer 53

well you got the critical point already right?

Answer 54

yup

Answer 55

the critical point for f(x)

Answer 56

yes

Answer 57

the other two points are the inflection points

Answer 58

to find them, you plug in x = 0 and x = 4 into f(x)

Answer 59

I got 0,-5 but it is not on the graph

Answer 60

it should be (0,5)

Answer 61

not sure how you got -5

Answer 62

I mean positive 5

Answer 63

I plugged 6.1 in and got neg 523

Answer 64

-523.8639, so close enough I guess

Answer 65

the inflection points I am confused with

Answer 66

well you found one and that was (0,5)

Answer 67

the other is found when you plug in x = 4

Answer 68

is it ok if it is not on my graph?

Answer 69

oh wait nvm

Answer 70

4,-315

Answer 71

and the range?

Answer 72

thanks for the help

Answer 73

plug in x = 6.1 to get the smallest possible y value (ie the absolute min)

this will help find the range

Answer 74

-524.8,infinity?

Answer 75

523

Answer 76

more like -523.8639 actually, but you get the idea

Answer 77

so the range is [-523.8639, infinity)

Answer 78

I guess -523.86 is more accurate

Answer 79

thanks a bunch man. why did my teacher give me such a tough example?

Answer 80

haha

Answer 81

that's what some teachers do

Answer 82

I really appreciate the patience.

Answer 83

and that's a good thing sometimes, you want your teacher to make you work hard

Answer 84

yw

Answer 85

should I put a place marker near 0,8?

Answer 86

what do you mean

Answer 87

the graph appears to be going around there or should I ignor it

Answer 88

not sure what you mean

Answer 89

the right half of the graph crosses x=8 something in the drawing

Answer 90

I might be over analyzing this

Answer 91

what about it?

Answer 92

should I put a point there or leave it alone

Answer 93

hmm it doesn't explicitly say "find the roots of f(x)", so I would say no

Answer 94

don't worry about it

Answer 95

ok haha perfect. Thanks a bunch!!!