Find an equation of the line tangent to the graph of

We just finished up limits and functions and then I log on to my webwork to see this stuff and I have no idea how to do it. If someone could explain it for me that would be great.
Find an equation of the line tangent to the graph of y=1-8x-3x^2
at the point (-3,-2)
Answer: y=
Thanks!
Answer #1
Maybe this helps you…
http://www.wolframalpha.com/input/?i=y%3D1-8x-3x^2+
Answer #2
Nope that doesn’t really help me at all. Thanks for your efforts though.
Answer #3
kingcoke replied: Nope that doesn't really help me at all. Thanks for your efforts though.
Like I said… It might help…
Sorry mate, I’m not that good in math…
Cheers!
Answer #4
I got it figured out answer was
y=10x+28
Answer #5
http://www.math.hmc.edu/calculus/tutorials/tangent_line/
y = 1 – 8x – 3x^2
y’ = -8 – 6x
x=-3 ==> y’ = -8 – 6(-3) = 10
y-(-2) = 10 (x-(-3))
y = 10x + 28 ——-> the equation of the tangent line
Answer #6
Hey d1ckh34dz I just downloaded your iatkos haha.
thanks for the breakdown.