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# Core 1 Differentiation question watch

1. 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.
2. 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.
3. (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 ...

The next equation is

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

Now solve these two simultaneously (once you've fixed your first question).
5. (Original post by GUMI)
Seriously
6. (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
7. (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.
8. (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 ...

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
9. (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.
10. (Original post by Zacken)
Yes, that's correct.
Thank you
11. (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
12. (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.

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