7. a. the curve with the equation ( y^{2}=x^{3}+3x^{2} ) is called the tschirnhausen cubic. find an equation…

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)).