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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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)

Mathematics
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You will need the first derivative. Go!
Is it 2x
\(\large\rm y'(x)=2x\) good :)

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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 \(\large\rm m\).
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.
\[\large\rm y'(2)=2(2)=m\]Ok with that part? :o
Yup, that makes sense
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.
Thanks so much!!

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