## anonymous 4 years ago 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.

1. anonymous

what is the equation first?

2. anonymous

first find the second coordinate by replacing x by -3

3. Mathhelp346

the equation would be 3 -3^2-8

4. anonymous

y=14

5. anonymous

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

6. anonymous

y=mx+b

7. anonymous

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

8. Mertsj

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

9. anonymous

yeah but its not coming out right

10. Mertsj

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

11. anonymous

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

12. Mertsj

Well plug in y = -18x-35

13. Mertsj

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

14. anonymous

I know but the website wasnt taking that answer

15. Mertsj

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

16. anonymous

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

17. anonymous

thanx

18. Mertsj

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

19. Mertsj

yw

20. anonymous

what a waste of 1 1/2 hours