Question

1. a. the curve with the 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?

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 solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx}=\frac{3x^{2}+6x}{2y}\).

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 \(y - y_{1}=m(x - x_{1})\) to find the tangent line equation

Here \(x_{1}=1,y_{1}=-2\) and \(m =-\frac{9}{4}\).
\(y+2=-\frac{9}{4}(x - 1)\).
Multiply through by \(4\): \(4y+8=-9x + 9\).
Rearrange to get \(9x+4y-1 = 0\).

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 \(x = 0\), from \(y^{2}=x^{3}+3x^{2}\), we have \(y^{2}=0\), so \(y = 0\).
When \(x=-2\), \(y^{2}=(-2)^{3}+3(-2)^{2}=-8 + 12 = 4\), so \(y=\pm2\).

Answer:

a. The equation of the tangent line is \(9x + 4y-1=0\).
b. The points with horizontal tangents are \((0,0),(-2,2),(-2,-2)\).