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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.

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  1. anonymous
    • 4 years ago
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    what is the equation first?

  2. anonymous
    • 4 years ago
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    first find the second coordinate by replacing x by -3

  3. Mathhelp346
    • 4 years ago
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    the equation would be 3 -3^2-8

  4. anonymous
    • 4 years ago
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    y=14

  5. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    y=mx+b

  7. anonymous
    • 4 years ago
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    \[y-19=-18(x+3)\] etc

  8. Mertsj
    • 4 years ago
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    Well imagine that. The same answer as when I helped you with the same problem.

  9. anonymous
    • 4 years ago
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    yeah but its not coming out right

  10. Mertsj
    • 4 years ago
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    What do you mean it's not coming out right?

  11. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    Well plug in y = -18x-35

  13. Mertsj
    • 4 years ago
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    You told me before that your book said the answer was y = -18x+b

  14. anonymous
    • 4 years ago
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    I know but the website wasnt taking that answer

  15. Mertsj
    • 4 years ago
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    Did you put in y = -18x-35 yet?

  16. anonymous
    • 4 years ago
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    but the website just took that answer so its right and the book didnt want me to simplify

  17. anonymous
    • 4 years ago
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    thanx

  18. Mertsj
    • 4 years ago
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    Well i think I told you that was the answer 1 1/2 hours ago.

  19. Mertsj
    • 4 years ago
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    yw

  20. anonymous
    • 4 years ago
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    what a waste of 1 1/2 hours

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