smart people good at math come

i need help

https://imgur.com/a/nbc2BRc

ok i believe the answer is this https://math.stackexchange.com/questions/3982545/line-l-is-tangent-to-fx-x42x3%E2%88%9211x2%E2%88%9213x35-at-x-p-and-x-q-find-the
thx everyone that helped, i dont get why this is true tho, 𝑓(𝑥)−𝐿=(𝑥−𝑝)^2(𝑥−𝑞)^2, but ill try to figure it out.

how much time you got? and do you want a full solved solution ?

the answer isn't going to be pretty

iriXium [#2]

how much time you got? and do you want a full solved solution ?

https://imgur.com/a/trGtXHz
i got this far but idk how to go from here, got the 2 system of equations but :v

oiiink [#4]

https://imgur.com/a/trGtXHz
i got this far but idk how to go from here, got the 2 system of equations but :v

give me 10 mins

Hynix [#3]

the answer isn't going to be pretty

yea i had a feeling 😭

iriXium [#5]

give me 10 mins

👍

This is just derivatives right?

oiiink [#6]

yea i had a feeling 😭

1. find derivative of f(x)
2. set up the tangent line equation at x=p and x=q
3. solve for points p and q
4. plug in slope and point to find equation of L

approx: y = -25.1x + 39.1, might have messed up somewhere but if u need an exact answer I can provide that as well

Serath [#8]

This is just derivatives right?

i think so

Hynix [#9]

1. find derivative of f(x)
2. set up the tangent line equation at x=p and x=q
3. solve for points p and q
4. plug in slope and point to find equation of L

approx: y = -25.1x + 39.1, might have messed up somewhere but if u need an exact answer I can provide that as well

i got the system of equation (#4) but idk how to solve for p and q from there :\

Smart people doesnt really do other people homework :c

oiiink [#7]

👍

waiiit

oiiink [#10]

i think so

just google this question

oiiink [#10]

i think so

Here have fun I even googled it for you:
https://math.stackexchange.com/questions/3982545/line-l-is-tangent-to-fx-x42x3%E2%88%9211x2%E2%88%9213x35-at-x-p-and-x-q-find-the

Anguibok [#12]

Smart people doesnt really do other people homework :c

Let them think they are smart. This is their chance.

Serath [#15]

Here have fun I even googled it for you:
https://math.stackexchange.com/questions/3982545/line-l-is-tangent-to-fx-x42x3%E2%88%9211x2%E2%88%9213x35-at-x-p-and-x-q-find-the

oh wow thx lol

Anguibok [#12]

Smart people doesnt really do other people homework :c

this is vlr where people love to play games with others
most asians are actually good in math even if they dont want to lol

I'm good at math but I haven't learned derivatives yet 💀

you are in college or school?

eldenlordkimmy [#20]

you are in college or school?

school

im not sure but i think you take derivative and find roots to find points at which slope of tangent are equal and maybe form line eq from that please correct me if i am wrong

Ask chatgpt bro that shit knows everything

Tryeue2 [#22]

im not sure but i think you take derivative and find roots to find points at which slope of tangent are equal and maybe form line eq from that please correct me if i am wrong

the derivative is 3rd degree so itll be hard to find the roots, i made 2 line eqations using p and q but i think its a bit hard to solve it, u have to realize that 𝑓(𝑥)−𝐿=(𝑥−𝑝)^2(𝑥−𝑞)^2 and solve from there

UvuvwevwevweOnyetenyevwe [#23]

Ask chatgpt bro that shit knows everything

I asked

To solve this problem, we'll follow these steps:

1. Find the derivative of ( f(x) ).
2. Determine the values of ( p ) and ( q ) where the tangent line is the same at both points.
3. Use the derivative to find the slope of the tangent line.
4. Use the points and the slope to find the equation of the tangent line.

Given:
[ f(x) = x^4 + 2x^3 - 11x^2 - 13x + 35 ]

1. Calculate the derivative ( f'(x) ):
   [ f'(x) = 4x^3 + 6x^2 - 22x - 13 ]

2. Set the derivative equal to zero to find the critical points (which might be our ( p ) and ( q )):
   [ 4x^3 + 6x^2 - 22x - 13 = 0 ]

Let's solve this cubic equation to find the roots ( p ) and ( q ). This can be done using numerical methods or by solving it analytically. For the sake of simplicity, we will use a symbolic approach to find the exact roots.

from sympy import symbols, solve

x = symbols('x')
f_prime = 4*x**3 + 6*x**2 - 22*x - 13
roots = solve(f_prime, x)
roots

Running this in a Python environment with symbolic computation capabilities (like SymPy) will give us the roots.

3. Suppose we find the roots to be ( p ) and ( q ). These are the ( x )-values where the tangent line is the same. Now we need to find the slope of the tangent line at these points. The slope ( m ) is given by ( f'(p) ).

4. Find the equation of the tangent line. Using the point-slope form of the line equation:
   [ y - f(p) = m(x - p) ]
   where ( m = f'(p) ).

Since the tangent line is the same at both ( x = p ) and ( x = q ), it implies that ( f'(p) = f'(q) = m ).

Finally, let’s find the roots and the equation of the tangent line step by step.

Step-by-Step Solution:

1. Derivative:
   [ f'(x) = 4x^3 + 6x^2 - 22x - 13 ]

2. Find roots ( p ) and ( q ):
   Solve ( 4x^3 + 6x^2 - 22x - 13 = 0 ).

3. Evaluate ( f(p) ) and ( f(q) ):
   [ f(p) = p^4 + 2p^3 - 11p^2 - 13p + 35 ]
   [ f(q) = q^4 + 2q^3 - 11q^2 - 13q + 35 ]

4. Find the slope ( m ):
   [ m = f'(p) = f'(q) ]

5. Equation of the tangent line:
   Using the point ( (p, f(p)) ):
   [ y - f(p) = m(x - p) ]

Finding the Roots

Let's use a symbolic solver to find the roots of ( f'(x) = 0 ):

from sympy import symbols, solve

x = symbols('x')
f_prime = 4*x**3 + 6*x**2 - 22*x - 13
roots = solve(f_prime, x)
roots

The roots are:
[ p = -2 ]
[ q = 1 ]

Tangent Line Calculation

1. Slope at ( p ) and ( q ):
   [ f'(-2) = 4(-2)^3 + 6(-2)^2 - 22(-2) - 13 = -32 + 24 + 44 - 13 = 23 ]

2. Calculate ( f(p) ) and ( f(q) ):
   [ f(-2) = (-2)^4 + 2(-2)^3 - 11(-2)^2 - 13(-2) + 35 = 16 - 16 - 44 + 26 + 35 = 17 ]

[ f(1) = 1^4 + 2(1)^3 - 11(1)^2 - 13(1) + 35 = 1 + 2 - 11 - 13 + 35 = 14 ]

3. Equation of the tangent line:
   [ y - f(-2) = f'(-2)(x + 2) ]
   [ y - 17 = 23(x + 2) ]
   [ y = 23x + 46 + 17 ]
   [ y = 23x + 63 ]

Therefore, the equation of the tangent line is:
[ y = 23x + 63 ]

Eragon_35 [#18]

this is vlr where people love to play games with others
most asians are actually good in math even if they dont want to lol

Then I'm fake asian XD

UvuvwevwevweOnyetenyevwe [#25]

I asked

To solve this problem, we'll follow these steps:

1. Find the derivative of ( f(x) ).
2. Determine the values of ( p ) and ( q ) where the tangent line is the same at both points.
3. Use the derivative to find the slope of the tangent line.
4. Use the points and the slope to find the equation of the tangent line.

Given:
[ f(x) = x^4 + 2x^3 - 11x^2 - 13x + 35 ]

1. Calculate the derivative ( f'(x) ):
   [ f'(x) = 4x^3 + 6x^2 - 22x - 13 ]

2. Set the derivative equal to zero to find the critical points (which might be our ( p ) and ( q )):
   [ 4x^3 + 6x^2 - 22x - 13 = 0 ]

Let's solve this cubic equation to find the roots ( p ) and ( q ). This can be done using numerical methods or by solving it analytically. For the sake of simplicity, we will use a symbolic approach to find the exact roots.

from sympy import symbols, solve

x = symbols('x')
f_prime = 4*x**3 + 6*x**2 - 22*x - 13
roots = solve(f_prime, x)
roots

Running this in a Python environment with symbolic computation capabilities (like SymPy) will give us the roots.

3. Suppose we find the roots to be ( p ) and ( q ). These are the ( x )-values where the tangent line is the same. Now we need to find the slope of the tangent line at these points. The slope ( m ) is given by ( f'(p) ).

4. Find the equation of the tangent line. Using the point-slope form of the line equation:
   [ y - f(p) = m(x - p) ]
   where ( m = f'(p) ).

Since the tangent line is the same at both ( x = p ) and ( x = q ), it implies that ( f'(p) = f'(q) = m ).

Finally, let’s find the roots and the equation of the tangent line step by step.

Step-by-Step Solution:

1. Derivative:
   [ f'(x) = 4x^3 + 6x^2 - 22x - 13 ]

2. Find roots ( p ) and ( q ):
   Solve ( 4x^3 + 6x^2 - 22x - 13 = 0 ).

3. Evaluate ( f(p) ) and ( f(q) ):
   [ f(p) = p^4 + 2p^3 - 11p^2 - 13p + 35 ]
   [ f(q) = q^4 + 2q^3 - 11q^2 - 13q + 35 ]

4. Find the slope ( m ):
   [ m = f'(p) = f'(q) ]

5. Equation of the tangent line:
   Using the point ( (p, f(p)) ):
   [ y - f(p) = m(x - p) ]

Finding the Roots

Let's use a symbolic solver to find the roots of ( f'(x) = 0 ):

from sympy import symbols, solve

x = symbols('x')
f_prime = 4*x**3 + 6*x**2 - 22*x - 13
roots = solve(f_prime, x)
roots

The roots are:
[ p = -2 ]
[ q = 1 ]

Tangent Line Calculation

1. Slope at ( p ) and ( q ):
   [ f'(-2) = 4(-2)^3 + 6(-2)^2 - 22(-2) - 13 = -32 + 24 + 44 - 13 = 23 ]

2. Calculate ( f(p) ) and ( f(q) ):
   [ f(-2) = (-2)^4 + 2(-2)^3 - 11(-2)^2 - 13(-2) + 35 = 16 - 16 - 44 + 26 + 35 = 17 ]

[ f(1) = 1^4 + 2(1)^3 - 11(1)^2 - 13(1) + 35 = 1 + 2 - 11 - 13 + 35 = 14 ]

3. Equation of the tangent line:
   [ y - f(-2) = f'(-2)(x + 2) ]
   [ y - 17 = 23(x + 2) ]
   [ y = 23x + 46 + 17 ]
   [ y = 23x + 63 ]

Therefore, the equation of the tangent line is:
[ y = 23x + 63 ]

i feel like that doesnt look right

UvuvwevwevweOnyetenyevwe [#25]

I asked

To solve this problem, we'll follow these steps:

1. Find the derivative of ( f(x) ).
2. Determine the values of ( p ) and ( q ) where the tangent line is the same at both points.
3. Use the derivative to find the slope of the tangent line.
4. Use the points and the slope to find the equation of the tangent line.

Given:
[ f(x) = x^4 + 2x^3 - 11x^2 - 13x + 35 ]

1. Calculate the derivative ( f'(x) ):
   [ f'(x) = 4x^3 + 6x^2 - 22x - 13 ]

2. Set the derivative equal to zero to find the critical points (which might be our ( p ) and ( q )):
   [ 4x^3 + 6x^2 - 22x - 13 = 0 ]

Let's solve this cubic equation to find the roots ( p ) and ( q ). This can be done using numerical methods or by solving it analytically. For the sake of simplicity, we will use a symbolic approach to find the exact roots.

from sympy import symbols, solve

x = symbols('x')
f_prime = 4*x**3 + 6*x**2 - 22*x - 13
roots = solve(f_prime, x)
roots

Running this in a Python environment with symbolic computation capabilities (like SymPy) will give us the roots.

3. Suppose we find the roots to be ( p ) and ( q ). These are the ( x )-values where the tangent line is the same. Now we need to find the slope of the tangent line at these points. The slope ( m ) is given by ( f'(p) ).

4. Find the equation of the tangent line. Using the point-slope form of the line equation:
   [ y - f(p) = m(x - p) ]
   where ( m = f'(p) ).

Since the tangent line is the same at both ( x = p ) and ( x = q ), it implies that ( f'(p) = f'(q) = m ).

Finally, let’s find the roots and the equation of the tangent line step by step.

Step-by-Step Solution:

1. Derivative:
   [ f'(x) = 4x^3 + 6x^2 - 22x - 13 ]

2. Find roots ( p ) and ( q ):
   Solve ( 4x^3 + 6x^2 - 22x - 13 = 0 ).

3. Evaluate ( f(p) ) and ( f(q) ):
   [ f(p) = p^4 + 2p^3 - 11p^2 - 13p + 35 ]
   [ f(q) = q^4 + 2q^3 - 11q^2 - 13q + 35 ]

4. Find the slope ( m ):
   [ m = f'(p) = f'(q) ]

5. Equation of the tangent line:
   Using the point ( (p, f(p)) ):
   [ y - f(p) = m(x - p) ]

Finding the Roots

Let's use a symbolic solver to find the roots of ( f'(x) = 0 ):

from sympy import symbols, solve

x = symbols('x')
f_prime = 4*x**3 + 6*x**2 - 22*x - 13
roots = solve(f_prime, x)
roots

The roots are:
[ p = -2 ]
[ q = 1 ]

Tangent Line Calculation

1. Slope at ( p ) and ( q ):
   [ f'(-2) = 4(-2)^3 + 6(-2)^2 - 22(-2) - 13 = -32 + 24 + 44 - 13 = 23 ]

2. Calculate ( f(p) ) and ( f(q) ):
   [ f(-2) = (-2)^4 + 2(-2)^3 - 11(-2)^2 - 13(-2) + 35 = 16 - 16 - 44 + 26 + 35 = 17 ]

[ f(1) = 1^4 + 2(1)^3 - 11(1)^2 - 13(1) + 35 = 1 + 2 - 11 - 13 + 35 = 14 ]

3. Equation of the tangent line:
   [ y - f(-2) = f'(-2)(x + 2) ]
   [ y - 17 = 23(x + 2) ]
   [ y = 23x + 46 + 17 ]
   [ y = 23x + 63 ]

Therefore, the equation of the tangent line is:
[ y = 23x + 63 ]

Beware that chatgpt doesn't guarantee you'll get the right answer

iriXium [#13]

waiiit

I think bro gave up 😭

one of the most important skills you have to learn as a student is knowing how to google shit

4 probably

cloudberry [#19]

I'm good at math but I haven't learned derivatives yet 💀

brotha what grade u in

cloudberry [#19]

I'm good at math but I haven't learned derivatives yet 💀

You'll learn them soon enough

oiiink [#27]

i feel like that doesnt look right

I don't think it looks right either
gives different answers every time

IonlywatchvcjXD [#28]

Beware that chatgpt doesn't guarantee you'll get the right answer

that is an understatement for maths xD
doing tutor work for ana1 at uni atm and the people who (imo) clearly just use chatgpt dont have an exactly good record of passing the sheets

UvuvwevwevweOnyetenyevwe [#34]

I don't think it looks right either
gives different answers every time

One thing that I learn while talking with my friends is you don't trust chatgpt with your math questions. They either understand what kind of question is it (e.g statistics, calculus) but always give you the wrong steps to do it, or chatgpt just straight bullshit their way through.

But I guess anything that's not math I think chatgpt can do it.

IonlywatchvcjXD [#26]

Then I'm fake asian XD

me too

Targu1n [#35]

that is an understatement for maths xD
doing tutor work for ana1 at uni atm and the people who (imo) clearly just use chatgpt dont have an exactly good record of passing the sheets

Lol I have never used chatgpt once but looking at the way my friends used chatgpt I kinda get that chatgpt can't do math cause the ai just straight bullshiting 🤣

IonlywatchvcjXD [#29]

I think bro gave up 😭

numbers are shitty like give a good polynomial ffs you are testing a concept not the ability of a student to calculate shit smh

IonlywatchvcjXD [#26]

Then I'm fake asian XD

most asians man not all lol

Tryeue2 [#39]

numbers are shitty like give a good polynomial ffs you are testing a concept not the ability of a student to calculate shit smh

Tbh I think it's better to be exposed with decimal numbers now, it's to make you get used to it 😅. Besides calculators are allowed there right?

oiiink [#24]

the derivative is 3rd degree so itll be hard to find the roots, i made 2 line eqations using p and q but i think its a bit hard to solve it, u have to realize that 𝑓(𝑥)−𝐿=(𝑥−𝑝)^2(𝑥−𝑞)^2 and solve from there

do you have any solution for it ? is it in terms of p and q?

IonlywatchvcjXD [#41]

Tbh I think it's better to be exposed with decimal numbers now, it's to make you get used to it 😅. Besides calculators are allowed there right?

if calculators are allowed you are chilling but why would you want to introduce useless fucking details to complicate questions. like make the questions hard instead.