# Thread: Tangent line to the curve

1. ## Tangent line to the curve

Find the equation for the to the given curve at the point where x = X₀

y=(x^2+3x-1)(2-x); X₀=1

I found the first derivative which came out to:

-x^2-3x+(2-x)(2x+3)+1
simplified to:
-3x^2-2x+7

then I set f( X₀)=1 to f( X₀)=(1)=2

Then I plugged it in my equation:

-3(1^2)-2(1)+7=2

I'm not sure if I'm doing it right and i'm not sure how to finish it. Thanks!

2. typo

3. Originally Posted by lisa1984wilson
Find the equation for the to the given curve at the point where x = X₀

y=(x^2+3x-1)(2-x); X₀=1

I found the first derivative which came out to:

-x^2-3x+(2-x)(2x+3)+1
simplified to:
-3x^2-2x+7

then I set f( X₀)=1 to f( X₀)=(1)=2

Then I plugged it in my equation:

-3(1^2)-2(1)+7=2

I'm not sure if I'm doing it right and i'm not sure how to finish it. Thanks!
y(1)=3
y'=-3x^2-2x+7
y'(1)=2
(y-3)=2(x-1)
this is the equation of the curve

4. I don't want to be a pain but how did you get the answers! I have a few others like this and I want to know how to do the rest.

5. $k=\frac{y-y_0}{x-x_0}$ and use the fact that the tangent $k=f'(x_0)$