# Easy tangent line problem but I don't know how to approach it

1. Sep 9, 2008

### Fiorella

Let f be the funtion given by f(x) = 4x^2 - x^3 , let L be the line y = 18 - 3x, where L is tangent to the graph of f.

Show that L is the tangent to the graph of y= f(x) at the point x = 3.

I'm equaling both graphs like this:

4x^2 - x^3 = 18 - 3x

and then I isolate everything to the right:

0 = x^3 - 4x^2 - 3x + 18

But I don't know how to factor this...am I in the right track to prove that this line is tangent to the function?

2. Sep 9, 2008

### Dick

Since you KNOW L is a tangent line, it's a bit easier to equate the derivatives. That just gives you a quadratic equation. Once you do that you still have to check f(x)=y for the two solutions. On the other hand you can factor that expression if you work at it. Review factoring. The only possible factors are (x-a) where a is a divisor of 18. And if (x-a) is a factor then substituting a into the polynomial gives you zero. Once you have a factor divide it out to reduce the degree. And don't forget to check y'=f'(x) for each root. Otherwise it's only a crossing point, not a tangent.

Hint: (x-3) is a factor.

3. Sep 9, 2008

### Fiorella

Ok Dick so I did synthetic division and I got:

x^2 - x - 6 (x - 3)

(x + 2) (x - 3) (x - 3)

x = -2 x=3 -----> so this proves that the line L is tangent to f(x) at x = 3?

I know somewhere I have to use derivatives to prove it, but I don't know how :(

Last edited: Sep 9, 2008
4. Sep 9, 2008

### Dick

That doesn't prove they are tangent. It proves the line intersects the curve at x=3. To show they are tangent you also need to show they have the same slope. Show that their derivatives are also equal at x=3.