Question

math help

Answer 1

@Vocaloid

Answer 2

Question?

Answer 3

https://s1.postimg.org/950bjtkonz/Capture.png

Answer 4

The ones you selected for 1 are correct
I'm not sure if option D is an answer too because
\((x^2+x)^2\) = \(x^4+2x^3+x^2\) and the leading is a 4th degree

Answer 5

And for #2 it is option B)

Answer 6

https://s1.postimg.org/96teghhp5r/Capture.png
number 4 a?

Answer 7

I am not sure about #3 you can ask vocaloid that

Answer 8

@Vocaloid

Answer 9

only look at the graph where x is between -2 and 3

what's the highest value the function obtains?

Answer 10

3?

Answer 11

is mehek correct on the other answers

Answer 12

make a tick mark where x = -2
make another tick mark where x = 3

between those two values, what's the highest y-value that the graph touches?

Answer 13

2

Answer 14

is the closest

Answer 15

good so 2 is your max

I think Mehek is right for the other answers so far

Answer 16

kinda stumped on 4 tbh, let me think

Answer 17

i'm getting better at pre cal lol trying to solve some on my own

Answer 18

my pre-calc is kind of rusty, but I believe since there are 2 local maxima there are at least 2 "turns" which means a minimum degree of 3 [just considering those roots] then you would consider the complex roots (2+3i and 2 - 3i)

Answer 19

Mehek do you remember how do to this? ;_;

Answer 20

I haven't taken this :/ so nope

Answer 21

can y'all check these next few

Answer 22

https://s1.postimg.org/4sskrjt2un/Capture.png

Answer 23

am i correct and what is 5

Answer 24

sorry this is taking so long, but yes, your answers for 6 and 7 are right

Answer 25

anyway, for the intermediate value theorem, you would evaluate the function the left hand and right hand ends of the interval (for example, evaluate f(-5) and f(-4). now, treat f(-5) to f(-4) as a new interval, and if 0 is within that interval, then the original interval contains a 0

Answer 26

I think it's a little easier to explain with math than with English

Answer 27

-3,-2 one?

Answer 28

can you show me your work?

Answer 29

what did you get for f(-3) and f(-2)?

Answer 30

i'm graphing it honestly

Answer 31

you can also do it algebraically

Answer 32

-4,-3 appears to one too

Answer 33

(-3,-2) does not contain a 0
(-4,-3) does

Answer 34

-5,-4
-4,-3
1,2

Answer 35

i'm looking at the graph on mathway probably should try desmos

Answer 36

am i right so far?

Answer 37

you're missing [0,1] take a look at the graph I attached

Answer 38

other than that I think you're good

Answer 39

http://www.wolframalpha.com/input/?i=x%5E4%2B6x%5E3-x%5E2-30x%2B4

Answer 40

wolfram alpha is a very powerful mathematical software

Answer 41

i thought it was 0,1

Answer 42

I'm talking about the intervals not points

Answer 43

[0,1] means the interval x = 0 to x = 1

in between x = 0 and x = 1, there is one zero so [0,1] is an option

Answer 44

keep in mind the answer choices are intervals, not points

Answer 45

https://s1.postimg.org/9n9ibnzw9b/Capture.png
but 0,1 is still an answer right and check 8 and help with 9?

Answer 46

8's good

Answer 47

for 9 I am not sure about formatting, but you would just find the possible roots using the p/q formula, and then test each root using synthetic division

Answer 48

(the rational root theorem, that is)

Answer 49

how would you do it if there isn't a q for example no x=3

Answer 50

hint: what is the coefficient on the highest powered term?

Answer 51

1

Answer 52

so x=1

Answer 53

x is not 1, but you would use that 1 to find the possible values of q

Answer 54

remember, q is all the possible factors of the coefficient on the highest powered term

so what are the factors of 1?

Answer 55

+-2i and -1

Answer 56

just consider real numbers for now

Answer 57

the only factor of 1 is 1

so your possible values of q are +1 and -1 (or written as +- 1)

Answer 58

what about the constant? (4)

what are the factors of 4?

Answer 59

2 and -2

Answer 60

keep going, you're missing 2 more

Answer 61

2i -2i

Answer 62

remember, we are only considering real roots for this (no i)

Answer 63

hint: 1 is a factor of every whole number, and if 1 is a factor of 4, what's also a factor of 4?

Answer 64

oh duh i'm going way ahead of myself

Answer 65

1,4

Answer 66

good so let's list out all the combinations

Answer 67

p = ± 4, ±2, ±1
q = ±1

so every possible combination of p/q can be tested as a root in synthetic division

Answer 68

so +4/-2, +1/1, etc.

Answer 69

so i need to divide 4/-2, 1/1 ?

Answer 70

yeah, you would divide all the possible combinations of p and q to find possible roots to test

Answer 71

so -2 and 1 ?

Answer 72

yes, those were two possible roots I used in the example

you would need to divide ALL combinations of p and q to find ALL the roots, though. it's more than just those two

Answer 73

-1/1=-1

Answer 74

good

keep going, there should be 6 possible roots

Answer 75

4/2=2

Answer 76

good

Answer 77

4/1=4 one more right?

Answer 78

good

Answer 79

2/1=2?

Answer 80

you already said this one ^^"

Answer 81

it's -4 to save you some time

Answer 82

oh oops

Answer 83

so our roots are +4, -4, +2, -2, +1, -1

so we can pick one root at a time to test

-1 is a good choice, so perform synthetic division with x = -1 (or, x + 1 = 0)

Answer 84

ok lol

Answer 85

there shouldn't be a remainder right i get 10

Answer 86

that would be for +1 as a root not -1 ^^

Answer 87

for x = -1 the number that goes outside the synthetic division symbol is -1 I believe?

Answer 88

or is it 1

Answer 89

x=1 i just calculated

Answer 90

ah, so x = 1 is not a root

try x = -1 as a root

Answer 91

https://s1.postimg.org/7k9f7wnk3j/Capture.png

Answer 92

awesome!

so now we have x^2 +4

do you think x^2 + 4 = 0 has any more roots?

Answer 93

that is, does x^2 + 4 = 0 have any real number solutions? (no i)

Answer 94

we said -2

Answer 95

hint: does (-2)^2 + 4 = 0?

Answer 96

i just said that lol

Answer 97

again, does (-2)^2 + 4 equal 0?

Answer 98

yes

Answer 99

hint: (-2)^2 = (-2)*(-2) = ?

Answer 100

-4

Answer 101

try one more time

Answer 102

negative * negative = positive

Answer 103

4

Answer 104

good

so (-2)^2+4 = 8, not 0

so -2 is not a root

Answer 105

oh crap duh i'm listening to an ll and doing pre cal

Answer 106

am done now

Answer 107

so do you think x^2 + 4 = 0 has any solutions?

Answer 108

good, it has no other solutions, so all possible roots have been found, making your problem complete

Answer 109

am almost done yay

Answer 110

https://s1.postimg.org/14tt3m5olr/Capture.png

Answer 111

10 should be right i need help with 11 mainly

Answer 112

I think you could do a piecewise function that excludes x = -2

Answer 113

and put (x-3) in the denominator to make that vertical asymptote

Answer 114

so for example i could say 6x^3-2x^2-x-3 for denominator?

Answer 115

I guess but you don't need to make it that complex

Answer 116

if you just put 1/(x-3) as your function it'll work fine

Answer 117

(it's your choice though)

Answer 118

anyway, after you have your function, you can write it in piecewise notation

{ f(x) = ______ from (-infinity,-2)
{ f(x) = _______ (same function) from (-2,infinity)

Answer 119

oh ok

Answer 120

so pick a function? for the numerator

Answer 121

actually come to think of it there's a better way

Answer 122

if you put (x+2) in your numerator AND the denominator this should give you a hole at x = -2

Answer 123

so x+2/x+2-x/3?

Answer 124

almost

Answer 125

(x+2)

Answer 126

whoops

Answer 127

(x+2) fraction bar (x+2)(x-3) should do it

Answer 128

how would i explain how i determined it

Answer 129

using the steps I have outlined above

Answer 130

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
if you put (x+2) in your numerator AND the denominator this should give you a hole at x = -2
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 131

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
and put (x-3) in the denominator to make that vertical asymptote
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 132

is 10 right? and 12? i'm about to solve 12

Answer 133

yeah I think 10's good

Answer 134

12 is a

Answer 135

https://s1.postimg.org/9ggmk51upb/Capture.png

Answer 136

1 more after this to check

Answer 137

12 is good

Answer 138

13?

Answer 139

yes good

Answer 140

for 14 find the least common denominator
convert both fractions to have the LCD
then just add

Answer 141

14 i have to solve

Answer 142

hint: to start, try factoring both denominators

then the LCD will be a bit more obvious

Answer 143

x x
-2 x

Answer 144

uh I'm not sure what this means ^^

Answer 145

x x
-4 1

Answer 146

try using parentheses to be more clear

Answer 147

x^2=(x*x) right

Answer 148

-2x=(-2*x)

Answer 149

if you have terms being added and subtracted you cannot factor each term individually, you have to consider the expression as a whole

Answer 150

for example, for (x^2-2x) what happens if you factor out x? make sure to keep parentheses in your answer

Answer 151

ypu have (x-2)

Answer 152

?

Answer 153

good, so we get (x)(x-2) as the new denominator

now try factoring (x^2-4) using difference of squares

Answer 154

x+2
x-2

Answer 155

good

so let's compare the left denominator and the right denominator

left: (x)(x-2)
right: (x+2)(x-2)

how can we get them both to be the same thing?

Answer 156

x^+2
x-2?

Answer 157

uh I'm not sure what you mean by that

Answer 158

x*x=x^2

Answer 159

I think you're a little off track ^^ let me help out

Answer 160

left: (x)(x-2)
right: (x+2)(x-2)

if we multiply the "right" by x and the "left" by (x+2) what do we get?

Answer 161

*** note: do not distribute any parentheses yet

Answer 162

x^4-2x^3-4x^2+8x

Answer 163

please do not distribute any parentheses yet

Answer 164

you should have gotten:

left: (x)(x-2)(x+2)
right: (x+2)(x-2)(x)

right?

Answer 165

oh ok so you wanted it exactly literally ok i'm over here trying to combine it

Answer 166

yeah, sorry if I wasn't clear ^_^"

Answer 167

piecewise function are not until next unit lol just looked for that other question it's ok

Answer 168

anyway, remember that, in order to keep the value of a fraction the same, you would multiply the numerator AND the denominator by the same value

Answer 169

so for the left-fraction, multiply the numerator and the denominator by (x+2)
and then for the right fraction, multiply the numerator and the denominator by x

Answer 170

if you do your math correctly you should have two fractions with the same denominator

then you can just add the numerators and put that over the new denominator if that makes sense

Answer 171

like:

(A/C) + (B/C) = (A+B)/C

Answer 172

so multiply with the x(x-2)(x+2) we have?

Answer 173

let me rephrase this ^^

Answer 174

or by the orginal equation

Answer 175

your original problem is

(x-4)/(x^2-2x) + (4/(x^2 - 4))

apply the multiplications to this ^

Answer 176

(you can use the factored form if that's easier)

Answer 177

x-4*(x+2)=-3x+2
x^2-2x*x+2=-x^2+2
4*x=4x
x^2-4*x=x^2-4x

Answer 178

check your math again

Answer 179

your denominators should be equal

Answer 180

use the factored form if that's simpler

Answer 181

(x-4)/ [x(x-2)] + (4/[x+2][x-2])

^ apply the multiplications to this

Answer 182

you should get x(x-2)(x+2) as your denominator

Answer 183

x+2/x^2+2? answer

Answer 184

mathway says the numerator is x+4?

Answer 185

try leaving things in factored form instead of expanding things out

Answer 186

at the very end, you can do the expansions

Answer 187

I wouldn't trust mathway, their software isn't good for calculations

Answer 188

also can we open a new question? this one is lagging a lot

Answer 189

I can transfer my work over

Answer 190

so x+2 for numerator is correct and sure

Answer 191

I don't think x + 2 is the correct numerator

Answer 192

im confused