Finding a tangent in a Cubic Function

[jahaddow]

Homework Statement

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

Homework Equations

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
Please help someone!

 

[lanedance]

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

 

[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)

 

[jahaddow]

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

 

[jahaddow]

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!

 

[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!

 

[jahaddow]

i can't think back that far, Please explain

 

[jahaddow]

ahh I know now, Point slope formula!