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#1
Question is down below
0
#2
A curve has equation
Y=x^3+px^2+qx-45
The curve passes through point R (2,3)
The gradient of the curve is 8
Find p and q
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#3
I got p= -12 and q=44 but not sure
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3 weeks ago
#4
I’m on my phone on a lunch break but as a starting point:

From the information given, you know that when x = 2, y = 3. You also know that when x = 3, y = 11. EDIT: no you don’t

You can put in those values and try to solve by trial and error, or by using simultaneous equations.
Last edited by MyMaths&chill; 3 weeks ago
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3 weeks ago
#5
(Original post by asifmahmoud)
A curve has equation
Y=x^3+px^2+qx-45
The curve passes through point R (2,3)
The gradient of the curve is 8
Find p and q
Dont you differentiate, then substitute x and make dy/dx=8. That makes one equation.
Then substitute coordinates into f(x) to get other equation, then solve simultaneously
Last edited by Chinemerem1; 3 weeks ago
0
3 weeks ago
#6
I got the same answer as you.
(Original post by asifmahmoud)
I got p= -12 and q=44 but not sure
0
3 weeks ago
#7
EDIT: I WAS WRONG DONT DO THIS

3 = 2^3 + p(2^2) + q(2) - 45
= 8 + 4p + 2q - 45
= 4p + 2q - 37

11 = 3^3 + p(3^2) + q(3) - 45
= 27 + 9p + 3q - 45
= 9p + 3q - 18

Therefore

4p + 2q = 40
9p + 3q = 29
Last edited by MyMaths&chill; 3 weeks ago
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3 weeks ago
#8
(Original post by MyMaths&chill)
I’m on my phone on a lunch break but as a starting point:

From the information given, you know that when x = 2, y = 3. You also know that when x = 3, y = 11.

You can put in those values and try to solve by trial and error, or by using simultaneous equations.
How do you know when x=3, y=11
0
#9
If so, when i solve the two equations simultaneously, i get p= -31/3 and q = 122/3 which doesn't seem to be the answer. I also don't know what's wrong with my answer either
(Original post by MyMaths&chill)
I’m on my phone on a lunch break but as a starting point:

From the information given, you know that when x = 2, y = 3. You also know that when x = 3, y = 11.

You can put in those values and try to solve by trial and error, or by using simultaneous equations.
0
3 weeks ago
#10
Oh **** yeah I did this wrong ignore what I said lol

Don’t mix lunch and maths, it doesn’t work
(Original post by Chinemerem1)
How do you know when x=3, y=11
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#11
If you calculate the gradient from those two points (2,3) and (3,11) you get gradient to be 8
(Original post by Chinemerem1)
How do you know when x=3, y=11
0
3 weeks ago
#12
(Original post by asifmahmoud)
If you calculate the gradient from those two points (2,3) and (3,11) you get gradient to be 8
That's for straight lines, this is a curve
0
3 weeks ago
#13
Yeah this is the correct approach.

Oops
(Original post by Chinemerem1)
Dont you differentiate, then substitute x and make dy/dx=8. That makes one equation.
Then substitute coordinates into f(x) to get other equation, then solve simultaneously
0
#14
You got me mate
(Original post by Chinemerem1)
That's for straight lines, this is a curve
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#15
So p= -12 and q= 44 is the right answer?
(Original post by Chinemerem1)
I got the same answer as you.
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3 weeks ago
#16
(Original post by asifmahmoud)
So p= -12 and q= 44 is the right answer?
That's what I got when I did it.
0
#17
(Original post by Chinemerem1)
That's what I got when I did it.
Alright. Thank you 😊
0
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