Question

(a) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point (1, 2). y = 9 4​x− 1 4​(b) At what points does this curve have horizontal tangents? (x, y) = (smaller y-value) (x, y) = (larger y-value)(c) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen

Answer 1

Answer:

  (a)  y = 9/4x -1/4

  (b)  (-2, -2), (-2, 2)

  (c)  see below

Step-by-step explanation:

(a) The derivative can be found from ...

  2y·y' = 3x² +6x

At (x, y) = (1, 2), this is ...

  4y' = 9

  y' = 9/4

so the equation of the tangent is ...

  y = (9/4)(x -1) +2

  y = (9/4)x -1/4

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(b) At y' = 0, we have ...

  0 = 3x² +6x = 3x(x +2)

This says y' = 0 at x=0 or x=-2. (x = 0 is an extraneous solution.)

At x = -2, we have ...

  y² = (-2)³ +3(-2)² = -8+12 = 4

  y = ±2

So, the horizontal tangent points are (x, y) = (-2, ±2).