# Coordinates at which 2 tangent lines of a function pass through a single point

1. Nov 11, 2007

### goaliemon89

Here is what the question asks:

Find the coordinate of all points in the graph of y=3e^x - x^3 at which the tangent line passes through the point (1,0).

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I am told that the 2 points are (-.872, -1.027) and (2.275, 13.657). Any help at all that you can give will be very much appreciated.

I was thinking of letting the 2 coordinates be (X1, Y1) and (X2, Y2), calculating the slope of the line through (1,0) and one of these points, and setting this slope equal to the derivative of y=3e^x - x^3 ......but I seem to be going no where.

Suggestions?

2. Nov 11, 2007

### JasonRox

Yes, the slope of the derivative would be the slope of the tangent at what point on the line?

3. Nov 11, 2007

### Galileo

What's the equation for the tangent line at the point (a,f(a))?

By the way. This final expression is nasty to solve. I think you'll need a computer to evaluate the answers numerically in the end.

4. Nov 11, 2007

### goaliemon89

Alright -- the final equation i get is 6e^x + 2x^3 - 3x^2 - 3xe^x = 0

The 2 answers that the book gives work in this equation......but how the hell would I solve this?

If I were to set 6e^x + 2x^3 - 3x^2 - 3xe^x equal to y in my calculator and graph it, is there a way for the calculator to tell you the x-intercepts?

EDIT - Never mind - I figure it out! Thanks for helping out.

Last edited: Nov 11, 2007
5. Nov 11, 2007

### JasonRox

Newton's Method?