# Stumbled upon a math problem

1. May 21, 2010

### jaycool1995

Hi everyone,
I have stumbled upon this maths problem, i have pondered it for a few days without any real work until today. The problem is, i can't seem to comprehend these questions at all... so any help would very much appreciated

Here is the math problem:

The curve C has equation y=k*x^3-x^2+x-5, where k is constant.

A) Find the derivative of the expression with respect to x

I got 3*k*x^2-2 x+1=y

The point A with x-coordinate -1/2 lies on c. The tangent c at a is parallel to the line with equation 2*y-7*x+1=0

Find:

B) the value of k

C) the value of the y-coordinate of A.

2. May 21, 2010

### Staff: Mentor

The derivative is correct, but you have mislabelled it as "y". You should have y' = 3kx^2 - 2x + 1 or dy/dx = 3kx^2 - 2x + 1.
Point A lies on C. The tangent to C at A is parallel to a certain line.

The derivative you found can be used to find the slope of the tangent line to C at any point. What's the slope of the tangent line (on C) at the point A? What's the slope of the line whose equation is given?

3. May 22, 2010

### jaycool1995

So, as far as i understand it, the slope of the line that is tangent to C at point A (-1/2). The second equation given is parallel to the first derivative of C when x=-1/2
My first thought was to do this:
$$3 k x^2-2 x+1=2 y+\frac{9}{2}$$
which is the same as

$$y'\left(-\frac{1}{2}\right) =2 y+\frac{9}{2}$$
which is equal to the line that is tangent to C. From this you can solve for k, which i got k=2. But i am struggling on how to find y, how could you use the other equation to find y, when the line is the same but rather only parallel. How could i find y in this case?
So wouldn't the equation look like this with the different y's:
$$\left(y_1\right)'\left(-\frac{1}{2}\right)\to \frac{3 k}{4}+2=2 y_2+\frac{9}{2}$$
or can we find y by going back to the original equation like this:
$$y=2 \left(-\frac{1}{2}\right)^3-\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)-5=-6$$
This is why it is confusing me lol i'll get there eventually ;)

Last edited: May 22, 2010