Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 32

a.) Find an equation of the tangent line to the curve $y^2 = x^3 + 3x^2$ which is also called Tschirnhausean Cubic, at the point $(1, -2)$

Solving for the slope $(m)$



$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (y^2) =& \frac{d}{dx} (x^3) + 3 \frac{d}{dx} (x^2)
\\
\\
2y \frac{dy}{dx} =& 3x^2 + (3)(2x)
\\
\\
2yy' =& 3x^2 + 6x
\\
\\
\frac{\cancel{2y}y'}{\cancel{2y}} =& \frac{3x^2 + 6x}{2y}
\\
\\
y' =& \frac{3x^2 + 6x}{2y}

\end{aligned}
\end{equation}
$


For $x = 1$ and $y = 2$, we obtain


$
\begin{equation}
\begin{aligned}

y' = m =& \frac{3(1)^2 + 6 (1)}{2(-2)}
\\
\\
m =& \frac{9}{-4}
\\
\\
m =& \frac{-9}{4}

\end{aligned}
\end{equation}
$


Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
\\
y - (-2) =& \frac{-9}{4} (x - 1)
\\
\\
y + 2 =& \frac{-9x + 9}{4}
\\
\\
y =& \frac{-9x + 9}{4} - 2
\\
\\
y =& \frac{-9x + 9 - 8}{4}
\\
\\
y =& \frac{-9x + 1}{4}
\qquad \text{Equation of the tangent line at $(1, -2)$}

\end{aligned}
\end{equation}
$


b.) Where does this curve have horizontal tangents?

Using the equation in part (a) to solve for $x$


$
\begin{equation}
\begin{aligned}

& y' = \frac{3x^2 + 6x}{2y} \qquad \qquad \text{where the slope is zero}
\\
\\
& 0 = \frac{3x^2 + 6x}{2y}
\\
\\
& 3x^2 + 6x = 0
\\
\\
& 3x(x + 2) = 0
\\
\\
& \begin{array}{c||c}
3x = 0 & x + 2 = 0 \\
x = 0 & x = -2
\end{array}

\end{aligned}
\end{equation}
$


Solving for $y$, when $x = 0$


$
\begin{equation}
\begin{aligned}

y^2 =& x^3 + 3x^2
\\
\\
y^2 =& (0)^3 + 3(0)^2
\\
\\
y^2 =& 0
\\
\\
y =& 0

\end{aligned}
\end{equation}
$


Solving for $y$, when $x = -2$


$
\begin{equation}
\begin{aligned}

y^2 =& (-2)^3 + 3(-2)^2
\\
\\
y^2 =& -8 + 12
\\
\\
y^2 =& 4
\\
\\
y =& \pm \sqrt{4}
\\
\\
y =& \pm 2

\end{aligned}
\end{equation}
$


The points where the curve have horizontal tangents are $(-2, -2), (-2, 2)$.

c.) Graph parts (a) and (b) using the curve and the tangent lines on a common screen.