# Derivatives and Tangents

1. Oct 1, 2007

### ThomasHW

1. The problem statement, all variables and given/known data
Find equations of all tangent lines to the graph of $$y=4x^{3}+5x-8$$
3. The attempt at a solution
I took the derivative of the equation, which was:

$$\acute{y}=12x^{2}+5$$

I remember having done these types of questions in high school, but I just can't remember, and I can't find any questions which are similar. Urg!

2. Oct 1, 2007

### Mindscrape

What does a derivative give you? You may be over-thinking this.

3. Oct 2, 2007

### ZioX

Essentially you look at the points (x,4x^3+5x-8) on the curve and for each point associate a tangent line. That is, solve (y-y(x_0))=y'(x_0)(x-x_0) for y.

4. Oct 2, 2007

### ThomasHW

It gives you the slope at any single point on a line.

I thought I should use the $$12x^{2}+5$$, and the point $$(1,-3)$$ in $$y=mx+b$$, to find a $$b$$ value, but when it's all said and done I don't get a tangent line. I get $$y=(12x^{2}+5)x-20$$ which just goes through the line.

BTW, they're looking for a total of two equations for the answer.

5. Oct 2, 2007

### HallsofIvy

Staff Emeritus
It gives you the slope of the tangent line! Which is exactly what you want.
And slope is a number not a formula in x. If you are looking for the slope of the tangent line at a point on the curve, you evaluate the derivative at the x value of that point.

Your original question said "Find equations of all tangent lines". I THINK you are now saying "find equations for all tangent lines to 4x3+ 5x- 8 that pass through (1, -3)" but you never told us about that last part!
One way to do that is not look at the derivative but look for solutions to the simultaneous equations y= m(x-1)+ 3 (any line through (1,3) can be written like that) and y= 4x3+ 5x- 8. For specific values of m that would give you a cubic for x which typically has three distinct answers. Look for the value of m so that equation has a double (or triple) root. That's how DesCartes found tangent lines "pre-calculus".

Last edited: Oct 2, 2007