Question

FIND THE equation of the tangent at x = 2
x^4 + x^3 + 1

Answer 1

i got 4x^3+ 3x^2

Answer 2

is that right

Answer 3

y=x^4+x^3+1
y' = 4x^3 + 3x^2
y'(2) = 4*2^3 + 3*2^2

Answer 4

and now just do:
y-y(2) = y'(2)(x-2)

Answer 5

wait i dont get that last part

Answer 6

y-y1=M(x-x1)

Answer 7

x1 = 2
y1 = y(2) = 2^4+2^3+1

Answer 8

why + 1

Answer 9

y = x^4 + x^3 + 1

Answer 10

y(2) = 2^4+2^3+1

Answer 11

oh oh kk

Answer 12

so is the right answer y=(4x^3+3x^2)(x)+1

Answer 13

wait i mean

Answer 14

No...
y-y(2) = y'(2)(x-2)
y - 2^4 - 2^3 - 1 = (4*2^3+3*2^2)(x-2

Answer 15

y = (4x^3+3x^2)(x)-63

Answer 16

is that right?

Answer 17

no, what i posted is. Just simplify it:
y - 2^4 - 2^3 - 1 = (4*2^3+3*2^2)(x-2)

Answer 18

but y is it y-y'
i dont understand the ewquation

Answer 19

i can do it i want to know y the equation is that way

Answer 20

it's y-y(2) = y'(2)(x-2)

Answer 21

right so if i was to simplify that

Answer 22

y(2) is y of 2 not two times y

Answer 23

i get y -25 = 4x^3+3x^2(x)-88

Answer 24

btw you leave y as y..

Answer 25

wat do u mean

Answer 26

y = 4x^3+3x^2(x)-63 ??

Answer 27

y - 2^4 - 2^3 - 1 = (4*2^3+3*2^2)(x-2)
Just simplify the numbers here...

Answer 28

ye thats wat i did

Answer 29

but should i isolate y?

Answer 30

then where are you getting 3x^2 and 4x^3 from......

Answer 31

cause thats wat y'(2) is

Answer 32

so y-y(2)=y'(2)(x-2)

Answer 33

y'(2) is 4*2^3 + 3*2^2..

Answer 34

y - 2^4 - 2^3 - 1 = (4*2^3+3*2^2)(x-2)
y - 25 = 44(x-2)

Answer 35

oo so then when the question asks for a derivative at a specific pooint then I would have to use what we did here but if it just says find the derivative of f(x) then i would put y = 4x^3+3x^2(x)

Answer 36

or no just y = 4x^3 + 3x^2

Answer 37

right?

Answer 38

Nooooo.... you're not following y(x) means u plug in whatever x is into y.. example:
y = 2x+1
x = 2
y(x) = 2*2+1 = 5

Answer 39

ye i get that im just saying if it says in general

Answer 40

to just din the derivative of some function and it is in rational format then i would use quotient rule to find the deravtive

Answer 41

find* the derivative

Answer 42

In general too. If you want to find an equation of the tangent line at x1 you find y'(x1). Then y(x1). Then you write:
y-y(x)=y'(x)(x-x1)

Answer 43

and that's it

Answer 44

right right i was talking about just a derivative not a tangent line

Answer 45

yea