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Core 1 Differentiation question

Hi, I'm doing this question for Core 1 revision but I can't find a and b.

Given that y=ax^2+bx-a^2 has gradient -4 at (-2,-13) find possible values for a and b.

I've differentiated the equation to get dy/dx=2ax+b-2a
Then I put the gradient in
-4=2a (-2)+b-2a
Now I'm at -4=-6a+b and I don't know what to do next or whether I've done something wrong.
You remove -a^2 , there's no -2a after. You are differentiating relative to x so you move (poor description) powers of x, if a value isn't a coefficient of x or a power of it then its removed.
Reply 2
Avatar for Zacken
Zacken
22
Original post by mydearestpotato
Hi, I'm doing this question for Core 1 revision but I can't find a and b.

Given that y=ax^2+bx-a^2 has gradient -4 at (-2,-13) find possible values for a and b.

I've differentiated the equation to get dy/dx=2ax+b-2a
Then I put the gradient in
-4=2a (-2)+b-2a
Now I'm at -4=-6a+b and I don't know what to do next or whether I've done something wrong.


Re-check your differentiation, the a^2 is just a constant. Otherwise it's like you saying:

Well, if I differentiate 4 = 2^2 then I get 2×22−12 \times 2^{2-1}...

The next equation is

-13 = a(-2) + b(-2) - a^2

Now solve these two simultaneously (once you've fixed your first question).
(edited 7 years ago)
Reply 3
Avatar for GUMI
GUMI
19
I would help you but idk what a gradient is
Original post by GUMI
I would help you but idk what a gradient is


Seriously
Original post by Vikingninja
You remove -a^2 , there's no -2a after. You are differentiating relative to x so you move (poor description) powers of x, if a value isn't a coefficient of x or a power of it then its removed.


Oh I see what you mean. I've tried again and I got dy/dx=2ax+b
Reply 6
Avatar for Zacken
Zacken
22
Original post by mydearestpotato
Oh I see what you mean. I've tried again and I got dy/dx=2ax+b


Yes, that's correct.
Original post by Zacken
Re-check your differentiation, the a^2 is just a constant. Otherwise it's like you saying:

Well, if I differentiate 4 = 2^2 then I get 2×22−12 \times 2^{2-1}...

The next equation is

-13 = a(-2) + b(-2) - a^2

Now solve these two simultaneously (once you've fixed your first question).



Okay so dy/dx=2ax+b

Then to make them equal would it be 2ax+b+4=-13+2a+2b+a^2
Reply 8
Avatar for Zacken
Zacken
22
Original post by mydearestpotato
Okay so dy/dx=2ax+b

Then to make them equal would it be 2ax+b+4=-13+2a+2b+a^2


Huh? Where did you get that from?!

Do you know what a simultaneous equation is and how to solve one?

You have two equations:

2a(-2) + b = -4

-13 = 2a - 2b - a^2

Now you can subsitute b = -4 - 2a(-2) into the second equation.
(edited 7 years ago)
Original post by Zacken
Yes, that's correct.

Thank you :smile:
Original post by Zacken
Huh? Where did you get that from?!

Do you know what a simultaneous equation is and how to solve one?

You have two equations:

2a(-2) + b = -4

-13 = 2a - 2b - a^2

Now you can subsitute b = -4 - 2a(-2) into the second equation.


I've solved it now, took me a while haha
I got a=-7, b=-32 and a=3,b=8
Reply 11
Avatar for Zacken
Zacken
22
Original post by mydearestpotato
I've solved it now, took me a while haha
I got a=-7, b=-32 and a=3,b=8


Well done. :biggrin:

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