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Anthonyn2121

  • one year ago

Find the slope of the curve at the given point P, and an equation of the tangent line at P y=x^2-3, P(2,1)

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  1. tkhunny
    • one year ago
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    You will need the first derivative. Go!

  2. Anthonyn2121
    • one year ago
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    Is it 2x

  3. zepdrix
    • one year ago
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    \(\large\rm y'(x)=2x\) good :)

  4. zepdrix
    • one year ago
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    A tangent line is just a straight line. So it'll have the form: \(\large\rm y=mx+b\) in slope-intercept form and \(\large\rm y-y_1=m(x-x_1)\) in point-slope form. The process of taking the derivative of the function is what gives us our \(\large\rm m\).

  5. zepdrix
    • one year ago
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    So at the point P(2,1), our x coordinate is 2. So we want to know the slope of our function (the derivative value), at x=2.

  6. zepdrix
    • one year ago
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    \[\large\rm y'(2)=2(2)=m\]Ok with that part? :o

  7. Anthonyn2121
    • one year ago
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    Yup, that makes sense

  8. zepdrix
    • one year ago
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    To get your final answer it would probably make more sense, at least for this problem, to go with point-slope form of a line. \[\large\rm y-y_1=m(x-x_1)\]Plug in your \(\large\rm m\), plug in your \(\large\rm P\), and bam you're done.

  9. Anthonyn2121
    • one year ago
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    Thanks so much!!

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