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Problem 34 Hard Difficulty

(a) The curve with equation $ y^2 = x^3 + 3x^2 $ is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point (1, -2).
(b) At what points does this curve have horizontal tangents?
(c) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen.

Answer

a. $y= \frac {-9x}{4}\times \frac{1}{4}$
b. Horizontal tangents occur at $(-2, \pm 2)$
c. SEE GRAPH

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CA

Catherine A.

October 27, 2020

Thought I needed a tutor to help with Calculus: Early Transcendentals, but this helps a lot more.

Video Transcript

So here we are given a curve with the equation Y squared equals three or x cubed plus three, expert. Okay, so with this in mind we want to implicitly differentiate um and find the equation of the tangent line at one negative two. So we see that when we calculate this we want to implicitly differentiate. So we'll have to y Y prime equals three X squared plus six X. And then we evaluate this at X equals one. We see that the slope why prime is going to be a negative 9/4. So we end up getting Y equals a negative nine worth acts plus 1/4. And we see that that is tangent to the line one negative two as expected. Um And by graphing these on the same screen, we're able to see that.