(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 4x− 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

The equation of the tangent line is: [tex]\mathbf{y = \frac 94x - \frac {1}4}[/tex]
The horizontal tangent points are [tex]\mathbf{(x, y) = (-2, \±2)}[/tex]

The equation is given as:

[tex]\mathbf{y^2 = x^3 + 3x^2}[/tex]

(a) Equation of the tangent line

Differentiate both sides of [tex]\mathbf{y^2 = x^3 + 3x^2}[/tex]

[tex]\mathbf{2y \cdot y' = 3x^2 + 6x}[/tex]

Make y' the subject

[tex]\mathbf{y' = \frac{3x^2 + 6x}{2y}}[/tex]

From the question, the points are given as:

[tex]\mathbf{(x,y) = (1,2)}[/tex]

Substitute these values in [tex]\mathbf{y' = \frac{3x^2 + 6x}{2y}}[/tex]

[tex]\mathbf{y' = \frac{3 \times 1^2 + 6 \times 1}{2 \times 2}}[/tex]

[tex]\mathbf{y' = \frac{3 + 6}{4}}[/tex]

[tex]\mathbf{y' = \frac{9}{4}}[/tex]

The equation of a tangent line is:

[tex]\mathbf{y = m(x - x_1) + y_1}[/tex]

Where:

[tex]\mathbf{m = y' = \frac{9}{4}}[/tex]

[tex]\mathbf{(x_1,y_1) = (1,2)}[/tex]

So, we have:

[tex]\mathbf{y = \frac 94(x - 1) + 2}[/tex]

Open bracket

[tex]\mathbf{y = \frac 94x - \frac 94 + 2}[/tex]

Take LCM

[tex]\mathbf{y = \frac 94x + \frac {-9 + 8}4}[/tex]

[tex]\mathbf{y = \frac 94x + \frac {-1}4}[/tex]

[tex]\mathbf{y = \frac 94x - \frac {1}4}[/tex]

Hence, the equation of the tangent line is: [tex]\mathbf{y = \frac 94x - \frac {1}4}[/tex]

(b) The points where the curve has horizontal tangents

In (a), we have:

[tex]\mathbf{2y \cdot y' = 3x^2 + 6x}[/tex]

Set y' to 0

[tex]\mathbf{2y \cdot 0 = 3x^2 + 6x}[/tex]

[tex]\mathbf{0 = 3x^2 + 6x}[/tex]

Expand

[tex]\mathbf{0 = 3x(x + 2)}[/tex]

Solve for x

[tex]\mathbf{3x = 0 \ or\ x + 2 = 0}[/tex]

[tex]\mathbf{x = 0 \ or\ x = -2}[/tex]

When x = 0,

[tex]\mathbf{y^2 = x^3 + 3x^2}[/tex]

[tex]\mathbf{y^2 = 0^3 + 3 \times 0^2}[/tex]

[tex]\mathbf{y = 0}[/tex]

When x = -2

[tex]\mathbf{y^2 = (-2)^3 + 3 \times (-2)^2}[/tex]

[tex]\mathbf{y^2 = 4}\\[/tex]

Take square roots

[tex]\mathbf{y = \pm2 }[/tex]

Hence, the horizontal tangent points are [tex]\mathbf{(x, y) = (-2, \±2)}[/tex]

(c) See attachment for graphs of (a) and (b)