tangent line help

[iron citrus]

Can someone pls walk me through the question. Idk which formula to use to find the equation of the tangent line.

[iron citrus]

what i did so far

[glacial frost]

well linear functions have the property that their derivative is a constant

so a tangent line at x = 1 is basically identical

to the linear function itself

the normal line's slope can be determined by doing this

-1/f'(x) = -1/m

snooze button

wake up

[iron citrus]

wait, let me try it rq

[iron citrus]

I dont get this

wdym by -1/m?

[glacial frost]

-1 over slope of tangent line

what is even a normal line

it is one line that intersects the tangent line perpendicular

[iron citrus]

yea

[glacial frost]

-1/m where m is the tangent's slope is basically the slope of the normal

here it refers as f' the derivative

but it's the same

[iron citrus]

the first step is finding f’(x)?

[glacial frost]

yea but you can read it

it's a linear function

the slope can be read

whether I calculate f'(x) which is 3

or read off from f(x) = 3x-4 that the slope is 3 doesnt matter because it's a linear function

[iron citrus]

ok f’(x) is 3

[glacial frost]

so knowing that what is the slope of the normal line

use the formula

[iron citrus]

1/3?

[glacial frost]

almost

[iron citrus]

wait is x1 =1?

[glacial frost]

ye

[iron citrus]

what do i with -1?

[glacial frost]

it's trivial doesn't matter

f'(x) = 3 for all x

henc

f'(-1) = 3

too

because it's linear

the slope is constant

it's -1/3

you forgot the minus

now you wanna make sure that the normal line passed through the tangent at (-1, f(-1))

[iron citrus]

how do u know its constant is it because the point is at x=1?

[glacial frost]

what is the derivative of f

it's f'(x) = 3

what is 3?

a constant

[iron citrus]

yes

ooooo

[glacial frost]

which makes sense

[iron citrus]

i unde

Understand

[glacial frost]

a linear function has everywhere the same slope

trivially

the tangent line at x = -1 is the function itself

think about it

identical

you are basically drawing a tangent line on a line

the outcome is the same

[iron citrus]

Got it

[glacial frost]

the are both on each other

[iron citrus]

Is y1=-1?

[glacial frost]

no

y is the function value of x = -1

f(x1)

is y

or basically

f(-1)

[iron citrus]

Is the equation of the tangent line
Y-5=3(x-1) ?

[glacial frost]

😄

yea

if oyu solve for y

[iron citrus]

REALLY?!

[glacial frost]

you will see they are identical

do + 5

on both sides

[iron citrus]

why?

[glacial frost]

no wait

f(-1) = -7 not -5

y-(y1)

y-(-7)

[iron citrus]

Wait

[iron citrus]

Yup i understand it

[glacial frost]

cool

[iron citrus]

so then the normal line is
Y-(-7) =-1/3 (x-1)

[glacial frost]

cool

[iron citrus]

TYSM

.solved

[fallow grove]

Solved

Post marked as solved by @iron citrus.

Use .unsolved if this was a mistake.

[glacial frost]

nope

Y-(-7) =-1/3 (x+1)

plus

[iron citrus]

.unsolved

[fallow grove]

Unsolved

Post marked as unsolved by @iron citrus.

Use .solved to mark as solved.

[glacial frost]

what's snoozing this time

[iron citrus]

Its not -7

[glacial frost]

3(-1)-4 = -3 -4 = -7

[iron citrus]

Its f(1)

not f(-1)

Cause the point is at x=1

[glacial frost]

oh

well then the work was for x = -1 lol

try it for x = 1

let's see if you understood something

[glacial frost]

here is everything you need

[glacial frost]

,w plot y -(-1) = 3(x-1) and y-(-1) =-(1/3)(x-1)

[gleaming notch]

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Requested by adoniszgks

[glacial frost]

yea seems good

also

[iron citrus]

?

[glacial frost]

y -(-1) = 3(x-1)

-> y + 1 = 3x -3

->y = 3x-4 (original line)

[glacial frost]

?

the point is (1, f(1))

or (1, -1)

[iron citrus]

ooh so if u plug it back in the equation is should be equal to the original

[iron citrus]

So my first step was useful

.solved

[fallow grove]

Solved

Post marked as solved by @iron citrus.

Use .unsolved if this was a mistake.

[glacial frost]

yes but ONLY if it's a linear function