Posted by **Samantha** on Wednesday, November 5, 2008 at 6:24am.

Find the line which passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point.

So I found the derivative which is 3x^2.

Let (a, a3) be the point of tangency.

3x^2 = (a3 - 1/4)/(a-0)

I'm not sure how to solve for a.

Yes, the point is (0,1/4) but it's not on the curve. It's on the tangent line. I'm not sure how to solve for a and a3 in that equation. My algebra is bad.

--------------------------------------

How do you do this question.

f(x) = (5x-1)/(2-3x), find f^(-1)(x). The answer is supposed to be (1+2x)/(5+3x).

--------------------------------------

And can you please explain this one to me.

Suppose that xy^(3)+2y^(2)+xy+2x^(3) = 0. Find the general expression for dy/dx. Also, find the value of dy/dx at the point where x = 1.

I have no idea how to differentiate that.

Thanks so much for the help!

## Answer this Question

## Related Questions

- Calculus Part 2 - y = tan(sqrtx) Find dy/dx. So I found it and I got the answer ...
- Calculus - Sketch a graph of the parabola y=x^2+3. On the same graph, plot the ...
- calculus - 1. Which of the following expressions is the definition of the ...
- calculus - A curve passes through the point (1,-11) and it's gradient at any ...
- calculus - the tangent line to the curve y=x^2 at the point (a,a^2) passes ...
- calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
- AP Calculus - Consider the curve given by x^2+4y^2=7+3xy a) Show that dy/dx=(3y-...
- calculus - Consider line segments which are tangent to a point on the right half...
- calculus - Find the equation of a straight line that passes through a point (1,5...
- AP AB Calculus - Linear approximation: Consider the curve defined by -8x^2 + 5xy...