Differentiation Tangent??

[plucky kayak]

PLS HELP ON THESE TWO QUESTIONS

<@&791435371564892232>

@strong trench 🙏

@torn anchor 🙏

[torn anchor]

first part we need to find the co-ordinates of Q

this will be when y=0 so u solve ur equation

[plucky kayak]

itll be y = 0 something rgight

yh

[torn anchor]

and we also need the derivative to find the gradient of the tangent

[plucky kayak]

whats the derivative?

[torn anchor]

wait

what yr u in?

[plucky kayak]

11

[torn anchor]

where is this question from?

[thin lark]

they do differentiation in y11

in FM

[plucky kayak]

a fm book my teacher game me

[torn anchor]

only igcse or fm

[torn anchor]

fine so u havent learnt it

[plucky kayak]

we do differentiation at fm gcse

[plucky kayak]

is there no other way?

[torn anchor]

there isnt

[plucky kayak]

bc she set is as homework

[torn anchor]

u need to differentiate y

u can do the second one tho

what type of line is the tangent at O going to be

[plucky kayak]

straight

[plucky kayak]

as in flat

[torn anchor]

ye horizontal

[plucky kayak]

yuh

[torn anchor]

so whats the equation of the tangent gonna be

its a horizontal line that goes thru O

[plucky kayak]

y = 0x + c

[torn anchor]

or

whats c gonna be

[plucky kayak]

0

[torn anchor]

ye so the line is?

[plucky kayak]

but thats for that specific point

[plucky kayak]

0 = 0+ 0

[thin lark]

i think this how u do it

[torn anchor]

well all horizontal lines have equation y=c right?

[plucky kayak]

oh i didnt know that

[torn anchor]

we just need the y co-ordinate to find the horizontal line through the origin

[plucky kayak]

where did u gte x = 3?

[torn anchor]

should be smth to remember
horizontal is y=c
and vertical is x=c

[thin lark]

basically

[plucky kayak]

ok thank you

[thin lark]

Q is a root of the equation

[torn anchor]

solve for x in 3x^2 - x^3 = 0

[thin lark]

so i just set the equation to 0 becase it crosses the x axes when y = 0

[plucky kayak]

i tried that

but i couldnt factorise it

[thin lark]

u factor out x^2

[plucky kayak]

yh i did that

[torn anchor]

well both terms got an x^2

so u got x^2(3-x)=0

now one of those needs to be 0

so either x^2 = 0 or 3-x = 0

which getx x=0 or x=3

[thin lark]

also you can tell that x^2 is a repeated root which when it hits x =0 it goes in the opposite direction

it hits x = 0 then goes straight back up

but for this question u dont gotta know what that is neccesarrily

[plucky kayak]

im confused on this part

[thin lark]

if you look at both terms

the common thing in them is x

[plucky kayak]

yes

[thin lark]

but you can take out a factor of x^2
because X^3 is just X^2 * x

the biggest thing you can take out of both terms is x^2

[plucky kayak]

yh so then u get x^2 ( 3 -x )

right?

[thin lark]

yes

[plucky kayak]

but nilo said one of those had to be 0

i dont understand why

[thin lark]

Q is the point on the equation such that it

y coordinate is 0

[torn anchor]

how do we solve quadratics?

after factorising

[plucky kayak]

we workout them out for 0

[torn anchor]

dyk why?

[plucky kayak]

so it equals 0

[torn anchor]

yeah

same idea here

if neither x^2 or 3-x is 0, then when we multiply them, it cant be 0

[thin lark]

im sure this is familiar to you

[plucky kayak]

if we multiply them what was thje point of factorising

[torn anchor]

as in thats the idea

[plucky kayak]

yes

[thin lark]

ok so you see the solutions

[torn anchor]

when we multiply 2 non-zero things, we never get 0

[thin lark]

they happen when y = 0

[torn anchor]

so at least one of those x^2 or 3-x is 0

so that x^2(3-x) is 0

[plucky kayak]

let me js take this all in

[plucky kayak]

ok i undertsnad that part

[thin lark]

ok if you know that you get the roots/solutions at y = 0,

then set y = 3x^2 - x^2 to
3x^2 - x^2 = 0

[plucky kayak]

yes

[thin lark]

then factorise

x^2 (3-x) = 0

[plucky kayak]

x^2 (3 -x ) = 0

[thin lark]

the x on the outside basically means that one solution is at 0 but we don't gotta worry about that

3-x = 0
x = 3
so one solution is at x =3

so Q has coordinates (3,0)

[plucky kayak]

( X x X ) ( 3 - x ) = 0

[plucky kayak]

ok yh one solution is at x = 3

ok i got x = 3

[thin lark]

ok now, the only way we can find an equation of the tangent in the form y = mx +c
is if we know the gradient of the tangent

[plucky kayak]

yh

[thin lark]

luckily, if we differentiate the function, what you are basically doing is finding the gradient function,

basically this gives u the gradient but in terms of x, so that if we subbed in any value of x, it would give us the gradient at that exact point

[plucky kayak]

yh i differentiated it and got 6x - 3x^2

[thin lark]

yeh ok

[plucky kayak]

so i sub in x = 3?

[thin lark]

now you can sub in x = 3

yes

[plucky kayak]

ok

[plucky kayak]

-9

so the gradient is -9

[thin lark]

now you can write it in the form y = mx +c

[plucky kayak]

y = 9x + c

[thin lark]

i can show u a faster way if u want

y - y1 = m(x-x1)

[plucky kayak]

sure

[thin lark]

where y1 is the y coordinate of the point
and x1 is the x coordinate

[plucky kayak]

oh yh i think i know that

[thin lark]

try use this one

[plucky kayak]

yep

[thin lark]

its way faster

[plucky kayak]

ok

[thin lark]

so,

y - 0 = -9(x-3)

and u can rearrange that

[plucky kayak]

to find what tho?

[thin lark]

the equation of the tangent

y = -9x + 27

[plucky kayak]

OHH

YHH

[thin lark]

if you get stuck doing that method u can resort to doing it normally but i prefer that way

[plucky kayak]

tysm i finally got the equation of the tangent

[thin lark]

also one more thing,

if any question talks about like the equation of the normal

its the same process but when you find the gradient of the tangent, e.g. 5
then the gradient of the normal would be -1/5
as they are perpendicular

[thin lark]

np

[plucky kayak]

yh negative reciprocal

[thin lark]

u got it

[plucky kayak]

so would the next question just be y = 1/9x + 27

[thin lark]

y - 0 = 1/9(x-3)

y intercept also changes

[plucky kayak]

oh

1/3

instead of 27

[thin lark]

it makes sense that the y intercept is different because it wouldn't make sense for a perpendicular line to have the same y intercept

[plucky kayak]

yh true

i dont know why i didnt think of that

[thin lark]

-1/3 but yh

[plucky kayak]

.

oh yes

i meant that

[plucky kayak]

tysm for the help

can i js ask

how did u know what to do

i was so stuck i didnt know what to do at all

[thin lark]

well most of it was in the question like it talked about point Q so i just found the coordinates of it first

[thin lark]

i mean im Y12 tho xd

[plucky kayak]

ohh

i just started doing differentiation this week

as soon as i got my head around that

my teacher hit us w these

[thin lark]

i think the hardest question you can get on something like this is where they give you two graphs that intersect at a point then tell you to find the equation of tangent/normal at that specific point

[thin lark]

yeh its quite challenging learning from scratch

but when u understand it, everything gets a bit easier

[plucky kayak]

yh it js takes ages for me to understand

but i actually enjpoy doing maths once i do

[thin lark]

at least you do gcse fm 😂
i didn't do it at gcse

challenging yourself

[plucky kayak]

yh i really didnt want to drop it

needed atleast something out of the ordinary

but hopefully i dont end up having to drop it further down

[thin lark]

if you put in the work i think you will find it fine

[mystic perch]

Ok, I'm trying see if I got the correct answer

for point Q I got (3,0)

Did a little differentiation and got the equation of the tangent of that line to be y=-9x+27

and lastly I got the equation of y=0

<@&791435371564892232>

[torn anchor]

ye looks good

[proper ice]

@plucky kayak do u need any more help

[plucky kayak]

Could you please break down the way they got x = 3