maths question please help?

Question

The curve C has the equation
Y=x² + ax + b

where a and b are constants

Given that the minimum point of C has coordinates (-2,5), find the values of a and b?

please help need to hand it on monday many thanks, be great if you could show your workings

Answer 1

this problem can be solved by differentiating da equation y = x^2 + ax + b.
dy/dx = 2x+a;
to obtain max or min point, in this case min.
dy/dx = 0; 2x+a=0; given (-2,5) as min point.
therefore x = -2 -> a = -2x
a = 4;

now replace a as 4 into the initial equation
y = x^2 + 4x + b;
it is known curve passes thru point (-2,5) (the min)

replace values of x nad y into the equation

5 = (-2)^2 +4(-2) +b
b = 9;

a = 4; b = 9;

pls check da arithmetic for errors.

Answer 2

Integral:

y' = 2x + a. Now substitute -2,5 for x,y:
5 = -4 + a ==> a = 9

Now back to original equation:

5 = 4 - 18 + b ==> b = 19

Answer 3

The curve is a parabola with one minimum value. Look up the formula for the minimum value, apply the coordinates and solve for a and b. I'm not doing it for you!

Answer 4

f(x) = x² + ax + b
f ` (x) = 2x + a
f "(x) = 2 is not required but indicates that turning point is a minimum turning point.
For turning point:-
2x + a = 0
x = - a / 2
f (- a / 2) = a² / 4 - a² / 2 + b
f (- a / 2) = - a² / 4 + b
Turning point is (- a / 2, b - a² / 4)

- a/ 2 = - 2
a = 4

b - a²/4 = 5
b - 4 = 5
b = 9

a = 4 , b = 9

Answer 5

y = x^2 + ax + b ................(1)
dy/dx = 2x + a ...................(2)

At a minimum, dy/dx = 0
Therefore from (2):
0 = 2x + a
x = -a/2
Because the minimum has an x co-ordinate of -2, it follows that:
-2 = -a/2
a = 4

Put x = -2 and y = 5 and a = 4 in (1):
5 = (-2)^2 + 4(-2) + b
5 = 4 - 8 + b
b = 5 - 4 + 8
b = 9.

Answer 6

Sorry I'm lost m8...but if you are in the UK you have until Tuesday...It's a bank holiday on Monday.

Answer 7

Its times like this I wish I had listened in Maths instead of dreaming about girls.....