Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 21

Suppose that $f(x) = 3x^2 - 5x$, find $f'(2)$ and use it to find an equation of the tangent line to the parabola $y = 3x^2 - 5x$ at the point $(2,2)$

Using the definition of the derivative of a function $f$ at a number $a$, denoted by $f'(a)$, is

$\qquad \displaystyle \qquad f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

We have,


$
\begin{equation}
\begin{aligned}

\qquad f'(a) =& \lim \limits_{h \to 0} \frac{3(a + h)^2 - 5 (a + h) - (3a^2 - 5a)}{h}
&& \text{Substitute $f(a + h)$ and $f(a)$}\\
\\
\qquad f'(a) =& \lim \limits_{h \to 0} \frac{\cancel{3a^2} + 6ah + 3h^2 - \cancel{5a} - 5h - \cancel{3a^2} + \cancel{5a}}{h}
&& \text{Expand and combine like terms}\\
\\
\qquad f'(a) =& \lim \limits_{h \to 0} \frac{3h^2 + 6ah - 5h}{h}
&& \text{Factor the numerator}\\
\\
\qquad f'(a) =& \lim \limits_{h \to 0} \frac{\cancel{h}(3h + 6a - 5)}{\cancel{h}}
&& \text{Cancel out like terms}\\
\\
\qquad f'(a) =& \lim \limits_{h \to 0} (3h + 6a - 5) = 3(0) + 6 a - 5
&& \text{Evaluate the limit}\\
\\
\qquad f'(2) =& 6a - 5
&& \text{Substitute the value of $(a)$}\\
\\
f'(2) =& 6(2) - 5
&& \text{Simplify}\\
\\


\end{aligned}
\end{equation}
$


$\qquad \fbox{$f'(2) = 7$} \qquad $ Slope of the tangent line at $(2,2)$

Using Point Slope Form where the tangent line $y = f(x)$ at $(a, f(a))$




$
\begin{equation}
\begin{aligned}

\qquad y - f(a) &= f'(a)(x - a)
&& \\
\\
\qquad y - 2 &= 7(x - 2)
&& \text{Substitute value of $a, f(a)$, and $f'(a)$}\\
\\
\qquad y &= 7x - 14 + 2
&& \text{Combine like terms}

\end{aligned}
\end{equation}
$


$\qquad \fbox{$y = 7x - 12$} \qquad$ Equation of the tangent line at $(2, 2)$