# Finding a tangent in a Cubic Function

1. Jun 5, 2010

1. The problem statement, all variables and given/known data
My cubic function is y=(x-6)(x-1)(x-9) or y=x^3-16x^2+69x-54
I need to find the tangent at the point x=2.5

2. Relevant equations

3. The attempt at a solution
All that I have managed to do is work out the y value for x=2.5, that is y=34.125

Last edited: Jun 5, 2010
2. Jun 5, 2010

### lanedance

the gradient is the derivative of your function, so differentiate, ie. find $\frac{dy}{dx}$
same as other post

3. Jun 5, 2010

### number0

Derive and plug in x = 2.5

In other words, solve for $$\frac{dy}{dx}$$, and plug in 2.5 for x:

y = x3-16x2+69x-54

$$\frac{dy}{dx}$$ = 3x2-32x+69

At x = 2.5, $$\frac{dy}{dx}$$, which is the tangent, equals

3(2.5)2-32(2.5)+69 = ?

I am too lazy to do the calculation, but here is the basic setup.

If you are looking for a tangent line, then use y'(2.5) as the slope for the slope

equation: y-y0 = f'(2.5)(x-x0)

Last edited: Jun 5, 2010
4. Jun 5, 2010

thankyou ever so much, if I have any more questions I will ask!!

5. Jun 5, 2010

Ok, so I now have the Gradient and the X and Y values of the tangent. How do I get the equation of the tangent.

ps. Realy sorry for all the trouble, you are a big help!

6. Jun 5, 2010

### hgfalling

Well, if you have the slope of a line and a point on the line, how do you find the equation of the line? Remember back to algebra!

7. Jun 5, 2010