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

a.) Determine the slope of the tangent line to the parabola $y = 4x - x^2$ at the point $(1, 3)$

$(i)\text{ Using the definition (Slope of the tangent line)}$

$\displaystyle \lim \limits_{x \to a} \frac{f(x) - f(a)} {x - a}$

Here, we have $a = 1$ and $f(x) = 4x - x^2$, so the slope is


$\begin{equation} \begin{aligned} \displaystyle m =& \lim \limits_{x \to 1} \frac{f(x) - f(a)}{x - 1}\\ \\ \displaystyle m =& \lim \limits_{x \to 1} \frac{4x - x^2 - [4(1) - (1)^2]}{x - 1} && \text{ Substituting value of $a$ and $x$}\\ \\ \displaystyle m =& \lim \limits_{x \to 1} \frac{4x - x^2 - 3}{x - 1} && \text{ Factor the numerator}\\ \\ \displaystyle m =& \lim \limits_{x \to 1} \frac{-1(x - 3) \cancel{(x - 1)}}{\cancel{x - 1}} && \text{ Cancel out like terms and simplify}\\ \\ \displaystyle m =& \lim \limits_{x \to 1} (-x+3) = -1+3= 2 && \text{ Evaluate the limit }\\ \end{aligned} \end{equation}$

Therefore,
The slope of the tangent line is $m=2$
$(ii)$ Using the equation

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

Let $f(x) = 4x - x^2$. So the slope of the tangent line at $(1, 3)$ is


$\begin{equation} \begin{aligned} \displaystyle m =& \lim \limits_{h \to 0} \frac{f(1 + h) - f(1)}{h}\\ \\ \displaystyle m =& \lim \limits_{h \to 0} \frac{4(1 + h) - ( 1 + h)^2 - [4(1) - (1)^2]}{h} && \text{ Subsitute value of $a$} \\ \\ \displaystyle m =& \lim \limits_{h \to 0} \frac{4 + 4h - ( 1 + 2h + h^2) - 3}{h} && \text{ Expand and simplify }\\ \\ \displaystyle m =& \lim \limits_{h \to 0} \frac{2h - h^2}{h} && \text{ Factor the numerator}\\ \\ \displaystyle m =& \lim \limits_{h \to 0} \frac{\cancel{h} ( 2 - h)}{\cancel{h}} && \text{ Cancel out like terms }\\ \\ \displaystyle m =& \lim \limits_{h \to 0} (2 - h) = 2 - 0 = 2 && \text{ Evaluate the limit}\\ \end{aligned} \end{equation}$

Therefore,
The slope of the tangent line is $m=2$
b.) Write an expression of the tangent line in part (a).

Using the point slope form


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

Therefore,
The equation of the tangent line at $(1,3)$ is $y = 2x + 1$
c.) Illustrate the graph of the parabola and the tangent line. As a check on your work, zoom in toward the point $(1,3)$ until
the parabola and the tangent line are indistinguishable.