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Find the slope-intercept equation of the tangent line to the graph of y=1/x^3 when x = ½.
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hi :) what have you tried?? slope of tangent at a point = derivative of the equation at that point!
\[ y = x^{-3} \] find dy/dx and evaluate at x=1/2 to get the slope then find the tangent point i.e. (x,y) on the curve y= x^-3 when x=1/2 you can write \[ y - y_0 = m(x-x_0) \] and then rewrite in the form y = mx +b
y = -48x+32? am I right? :) so sorry that i was unable to respond awhile ago my internet was malfunctioning :)))
@hartnn @phi yeah and my answer was incorrect at first because my derivative is incorrect. Thank you guys :)
oh and i thought it was raised to -8 :)))
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Yes, looks good. Here is a plot
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