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.