# Math Help - derivative tangents

1. ## derivative tangents

I'd be eternally greatful if someone could help me with these few problems...

1. The Curve C has the equation

y=x²+ax+b

Where a and b are constants.

Given that the minimum point of C has co-ordinates (-2, 5), Find the values of a and b.

2. Given that

dy/dx=2x³+1,

and that y=3 when x=0, find the value of y when x=2.

And finally

3. f(x)=4x-3x²-x³

(a) Fully factorise f(x)=4x-3x²-x³.

I hope someone can help! It needs to be in by tomorrow, and one person has already been thrown off the course as their work was not upto standard. I fear mine will be without this help! Thanks in advance

2. Originally Posted by Joelmalmsteen
1. The Curve C has the equation

y=x²+ax+b

Where a and b are constants.

Given that the minimum point of C has co-ordinates (-2, 5), Find the values of a and b.
The minimum point of the curve will be the x for which $y' = 0$. So:
$y'(x) = 2x + a = 0$
has the solution
$x = \frac{-a}{2}$

We know that the minimum point of the curve is at x = -2, so thus
$-2 = \frac{-a}{2}$

or a = 4.

So $y = x^2 + 4x + b$
has a point (-2, 5) on it.

Thus:
$5 = (-2)^2 + 4(-2) + b$

or
$b = 9$

Thus your curve is $y = x^2 + 4x + 9$.

-Dan

3. Originally Posted by Joelmalmsteen

3. f(x)=4x-3x²-x³

(a) Fully factorise f(x)=4x-3x²-x³.
$x(4-3x-x^2)$
$-x(x^2+3x-4)$
$-x(x+4)(x-1)$

4. Originally Posted by Joelmalmsteen
2. Given that

dy/dx=2x³+1,

and that y=3 when x=0, find the value of y when x=2.
$\frac{dy}{dx} = 2x^3 + 1$

So
$y = \int dx \, (2x^3 + 1) = \frac{1}{2}x^4 + x + C$

We know that the point (0,3) is on this solution curve, so
$3 = \frac{1}{2}(0)^4 + (0) + C$

Thus C = 3.
$y = \frac{1}{2}x^4 + x + 3$

What is y when x = 2?
$y = \frac{1}{2}(2)^4 + 2 + 3 = 8 + 2 + 3 = 13$

-Dan