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anonymous

  • one year ago

Find the equation of the tangent line to the curve at the given point. y = 5x − 2squarerootx, (1, 3)

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  1. Mertsj
    • one year ago
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    Can you find the first derivative of the function?

  2. Mertsj
    • one year ago
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    @ElfQueen Hello. Are you there?

  3. anonymous
    • one year ago
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    You may find the first derivative and then find f'(1) which will give you your slope, and then use the following equation \[y-y_1=m(x-x_1)\]

  4. anonymous
    • one year ago
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    @Mertsj Sorry. I was making myself dinner. I got 5 - 1over the squareroot x.

  5. anonymous
    • one year ago
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    That's correct \[f'(x) = 5-\frac{ 1 }{ \sqrt{x} }\] now find f'(1)

  6. anonymous
    • one year ago
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    4

  7. anonymous
    • one year ago
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    Yes! So our slope is 4, now we can use the point slope equation and we will have our equation

  8. anonymous
    • one year ago
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    4x-1

  9. anonymous
    • one year ago
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    I'm not sure what that means, we just have to use \[y-y_1=m(x-x_1)\] and we are given the point (x1,y1)

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