Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 9

a.) Determine the slope of the tangent to the curve $y = 3 + 4x ^2 - 2x^3$ at the point where $x = a$

Using the equation

$ \displaystyle m = \lim \limits_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Let $f(x) = 3 + 4x^2 - 2x^3$. So the slope of the tangent to the curve at the point where $x = a$


$
\begin{equation}
\begin{aligned}

\displaystyle m &= \lim \limits_{h \to 0} \frac{f(a + h) - f(a)}{h}\\
\\
\displaystyle m &= \lim \limits_{h \to 0} \frac{3 + 4 (a + h)^2 - 2 (a + h)^3 - (3 + 4 a^2 - 2a^3)}{h}
&& \text{ Substitute value of $a$}\\
\\
\displaystyle m &= \lim \limits_{h \to 0} \frac{3 + 4 (a^2 + 2ah + h^2) - 2 (a^3 + 3a^2 h + 3ah^2 + h^3) - 3 - 4a^2 + 2a^3}{h}
&& \text{Expand and simplify}\\
\\
\displaystyle m &= \lim \limits_{h \to 0} \frac{\cancel{3} + \cancel{4a^2} + 8ah + 4h^2 - \cancel{2a^3} - 6 a^2 h - 6ah^2 - 2h^3 - \cancel{3} - \cancel{4 a^2} + \cancel{2a^3}}{h}
&& \text{Combine like terms}\\
\\
\displaystyle m &= \lim \limits_{h \to 0} \frac{8ah + 4h^2 - 6a^2 h - 6ah^2 - 2h^3}{h}
&& \text{Factor the numerator}\\
\\
\displaystyle m &= \lim \limits_{h \to 0} \frac{\cancel{h} (8a + 4h - 6a^2 - 6ah - 2h^2)}{\cancel{h}}
&& \text{Cancel out like terms}\\
\\
\displaystyle m &= \lim \limits_{h \to 0} (8a + 4h - 6a^2 - 6ah - 2h^2) = 8a + 4(0) - 6a^2 - 6a(0) - 2(0)^2 = 8a-6a^2
&& \text{Evaluate the limit}\\
\end{aligned}
\end{equation}
$

Therefore,
The slope of the tangent line is $m = 8a - 6a^2$

b.) Determine the equations of the tangent lines at the points $(1, 5)$ and $(2, 3)$

Solving for the slope of the tangent line at $(1, 5)$

Using the equation of slope of the tangent in part (a), we have $a = 1$ So the slope is


$
\begin{equation}
\begin{aligned}

m =& 8a - 6a^2 && \\
\\
m =& 8(1) - 6(1)^2
&& \text{Substitute value of $a$ and simplify}\\
\\
m =& 2
&& \text{Slope of the tangent line at $(1, 5)$}

\end{aligned}
\end{equation}
$


Using point slope form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1) && \\
\\
y - 5 =& 2 (x - 1)
&& \text{Substitute value of $x, y$ and $m$}\\
\\
y =& 2x - 2 + 5
&& \text{Combine like terms}\\
\\
y =& 2x + 3
\end{aligned}
\end{equation}
$

Therefore,
The equation of the tangent line at $(1,5)$ is $ y = 2x + 3$
Solving for the slope of the tangent line at $(2, 3)$

Using the equation of slope of the tangent in part (a), we have $a = 2$ so the slope is


$
\begin{equation}
\begin{aligned}

m =& 8a - 6a^2\\
&& \\
m =& 8(2) - 6(2)^2
&& \text{Substitute value of $a$ and simplify}\\
\\
m =& -8
&& \text{Slope of the tangent line at $(2, 3)$}
\end{aligned}
\end{equation}
$


Using point slope form

$
\begin{equation}
\begin{aligned}

y - y-1 =& m(x - x_1)
&& \\
\\
y - 3 =& -8 (x - 2)
&& \text{Substitute value of $x, y$ and $m$}\\
\\
y =& -8x + 16 + 3
&& \text{Combine like terms}\\
\\
y =& -8x+19


\end{aligned}
\end{equation}
$


Therefore,
The slope of the tangent line at $(2,3)$ is $y = -8x+19$
c.) Draw the graph of the curve and both tangent lines on a common screen.