Question

a graphing calculator is recommended.
(a) the curve with equation ( y^{2}=x^{3}+3 x^{2} ) 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? (order your answers from smallest to largest ( x ), then from smallest to largest ( y ).)
( (x, y)=(quad) )
( (x, y)=(quad) )
(c) illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen.

Explanation:

Step1: Differentiate the equation implicitly

Differentiate \(y^{2}=x^{3}+3x^{2}\) with respect to \(x\).
Using the chain - rule \((y^{2})^\prime = 2y\frac{dy}{dx}\) and the power - rule \((x^{n})^\prime=nx^{n - 1}\), we get:
\(2y\frac{dy}{dx}=3x^{2}+6x\).
Then \(\frac{dy}{dx}=\frac{3x^{2}+6x}{2y}\).

Step2: Find the slope of the tangent line at the point \((1,2)\)

Substitute \(x = 1\) and \(y = 2\) into \(\frac{dy}{dx}\).
\(\frac{dy}{dx}\mid_{(x = 1,y = 2)}=\frac{3(1)^{2}+6(1)}{2(2)}=\frac{3 + 6}{4}=\frac{9}{4}\).

Step3: Use the point - slope form of a line \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=1,y_{1}=2,m=\frac{9}{4}\).
\(y - 2=\frac{9}{4}(x - 1)\).
Expand: \(y-2=\frac{9}{4}x-\frac{9}{4}\).
\(y=\frac{9}{4}x-\frac{9}{4}+2=\frac{9}{4}x-\frac{9}{4}+\frac{8}{4}=\frac{9}{4}x-\frac{1}{4}\).

Step4: Find the points with horizontal tangents

A horizontal tangent occurs when \(\frac{dy}{dx}=0\). Since \(\frac{dy}{dx}=\frac{3x^{2}+6x}{2y}\), set \(3x^{2}+6x = 0\).
Factor out \(3x\): \(3x(x + 2)=0\). So \(x=0\) or \(x=-2\).
When \(x = 0\), then \(y^{2}=0^{3}+3(0)^{2}=0\), so \(y = 0\).
When \(x=-2\), then \(y^{2}=(-2)^{3}+3(-2)^{2}=-8 + 12 = 4\), so \(y=\pm2\).

Answer:

(a) \(y=\frac{9}{4}x-\frac{1}{4}\)
(b) \((x,y)=(-2,-2)\), \((x,y)=(-2,2)\), \((x,y)=(0,0)\)