Let f(x) = 3 x^2 - 8.
Find an equation of the line tangent to the graph of f at the point where x = -3.

Question

Let f(x) = 3 x^2 - 8.
Find an equation of the line tangent to the graph of f at the point where x = -3.

anonymous · Answer

what is the equation first?

anonymous · Answer

first find the second coordinate by replacing x by -3

mathhelp346 · Answer

the equation would be 3 -3^2-8

anonymous · Answer

y=14

anonymous · Answer

i get 
$$y=3\times 9-8=27-8=19$$ so the point on your line is 
$$(-3,19)$$ then find the slope.  the derivative is 
$$y'=6x$$ and replacing x by -3 gives you 
$$-18$$ so you have a slope and a point, use point slope formula

anonymous · Answer

y=mx+b

anonymous · Answer

$$y-19=-18(x+3)$$ etc

mertsj · Answer

Well imagine that.  The same answer as when I helped you with the same problem.

anonymous · Answer

yeah but its not coming out right

mertsj · Answer

What do you mean it's not coming out right?

anonymous · Answer

I plugged in y=-18x+b and y-19=-18(x+3) and on the website Im supposed to submit is saying that it is incorrect

mertsj · Answer

Well plug in y = -18x-35

mertsj · Answer

You told me before that your book said the answer was y = -18x+b

anonymous · Answer

I know but the website wasnt taking that answer

mertsj · Answer

Did you put in y = -18x-35 yet?

anonymous · Answer

but the website just took that answer  so its right  and the book didnt want me to simplify

anonymous · Answer

thanx

mertsj · Answer

Well i think I told you that was the answer 1 1/2 hours ago.

mertsj · Answer

yw

anonymous · Answer

what a waste of 1 1/2 hours