Question

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 = (b) At what points does this curve have horizontal tangents? (x, y) = (smaller y-value) (x, y) = (larger y-value)?

Answer 1

The equation of the tangent line to the Tschirnhausen cubic curve at the point (1, 2) is y = (9/4)x + (7/4). The curve has horizontal tangents at the points (0, 0) and (-2, 8).

Step-by-step explanation:

The equation of the tangent line to the Tschirnhausen cubic curve at the point (1, 2) can be found by taking the derivative of the equation and substituting the x-coordinate of the given point in it.

Given curve equation: y² = x³ + 3x²

Taking the derivative with respect to x:

2yy' = 3x² + 6x

Substituting x = 1 and y = 2, we get:

2(2)y' = 3(1)² + 6(1)

4y' = 3 + 6

4y' = 9

y' = 9/4

So, the slope of the tangent line is 9/4.

Using the point-slope form of a line, we can write the equation of the tangent line as:

y - 2 = (9/4)(x - 1)

y = (9/4)x + (7/4)

The equation of the tangent line to the Tschirnhausen cubic curve at the point (1, 2) is y = (9/4)x + (7/4).

The curve will have a horizontal tangent when the slope is 0. Therefore, setting the derivative equal to 0 and solving for x, we get:

2yy' = 3x² + 6x

2y(0) = 3x² + 6x

3x² + 6x = 0

x(x + 2) = 0

x = 0 or x = -2

Substituting these values back into the equation of the curve, we get two points where the curve has horizontal tangents: (0, 0) and (-2, 8).