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DLS

  • 2 years ago

Find the equation to the tangent to the curve x^2+3y-3=0 which is parallel to the line y=4x-5

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  1. DLS
    • 2 years ago
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    \[\LARGE \frac{dy}{dx}=2x+3=4\] \[\LARGE x=-\frac{1}{2},y=\frac{11}{12}\]

  2. DLS
    • 2 years ago
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    \[\LARGE (y-\frac{11}{12})=4(x+\frac{1}{2})\]

  3. DLS
    • 2 years ago
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    my attempt^

  4. amistre64
    • 2 years ago
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    why x=-1/2 ?

  5. DLS
    • 2 years ago
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    oops 1/2

  6. amistre64
    • 2 years ago
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    other than that, your process is good

  7. DLS
    • 2 years ago
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    12y-48x+13=0 is what im getting :O

  8. amistre64
    • 2 years ago
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    y = -5/4 y+5/4 = 4(x-1/2) this is a line equation ... so if form does not matter this would suffice

  9. amistre64
    • 2 years ago
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    otherwise, algrbrate it to your hearts content

  10. DLS
    • 2 years ago
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    isnt y 11/12?

  11. amistre64
    • 2 years ago
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    .5^2 + 3(.5) - 3 .25 + 1.5 - 3 1.75 - 3 = -1.25 = -5/4

  12. amistre64
    • 2 years ago
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    since x = 1/2 and NOT -1/2 you have to reevaluate the y you found

  13. DLS
    • 2 years ago
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    arent we substituting x=1/2 in the equation of curve?

  14. amistre64
    • 2 years ago
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    yes, which is what i just did

  15. DLS
    • 2 years ago
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    okay! thanks! got it

  16. amistre64
    • 2 years ago
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    \[\frac14+\frac32-\frac31\] \[\frac14+\frac64-\frac{12}4\] \[\frac{7-12}4\]

  17. DLS
    • 2 years ago
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    @amistre64 sorry,it was 3Y and not 3X that is how I got 11/12...

  18. DLS
    • 2 years ago
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    it is x squared so -1/2 and 1/2 both are same that is why i didnt change the y component

  19. amistre64
    • 2 years ago
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    ohh then your derivative might need some looking at: x^2+3y-3=0 2x +3y' = 0 y' = -2x/3 = 4 x = 12/-2 = -6

  20. amistre64
    • 2 years ago
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    (6)^2+3y-3=0 33+3y=0, when y = -11

  21. amistre64
    • 2 years ago
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    y+11 = 4(x+6) would be the tangent line

  22. DLS
    • 2 years ago
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    (-6,-11)

  23. DLS
    • 2 years ago
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    CORRECT NOW! :D

  24. amistre64
    • 2 years ago
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    yay! ;)

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