Find the slope-intercept equation of the tangent line to the graph of y=1/x^3 when x = ½.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Find the slope-intercept equation of the tangent line to the graph of y=1/x^3 when x = ½.

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

hi :) what have you tried?? slope of tangent at a point = derivative of the equation at that point!
  • phi
\[ y = x^{-3} \] 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 :)))

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@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 :)))
  • phi
Yes, looks good. Here is a plot
1 Attachment

Not the answer you are looking for?

Search for more explanations.

Ask your own question