Using:

[latex]f^{\prime}\left(a\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(a+h\right)-f\left(a\right)}{h}[/latex]

Substitute [latex]f\left(a+h\right)={\left(a+h\right)}^{2}-4\left(a+h\right)[/latex] and [latex]f\left(a\right)={a}^{2}-4a[/latex].

[latex]\begin{align} {f}^{\prime }\left(a\right)&=\underset{h\to 0}{\mathrm{lim}}\frac{\left(a+h\right)\left(a+h\right)-4\left(a+h\right)-\left({a}^{2}-4a\right)}{h} \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{{a}^{2}+2ah+{h}^{2}-4a - 4h-{a}^{2}+4a}{h}&& \text{Remove parentheses}. \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{{a}^{2}+2ah+{h}^{2}-4a-4h-{a}^{2}+4a}{h}&& \text{Combine like terms}. \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{2ah+{h}^{2}-4h}{h} \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{\cancel{h}\left(2a+h - 4\right)}{\cancel{h}}&& \text{Factor out }h. \\ &=2a+0 - 4 \\ {f}^{\prime }\left(a\right)&=2a - 4&& \text{Evaluate the limit}. \\ {f}^{\prime }\left(3\right)&=2\left(3\right)-4=2 \end{align}[/latex]

Equation of tangent line at [latex]x=3:[/latex]

[latex]\begin{align}&y=f^{\prime}\left(a\right)\left(x-a\right)+f\left(a\right) \\ &y=f^{\prime}\left(3\right)\left(x - 3\right)+f\left(3\right) \\ &y=2\left(x - 3\right)+\left(-3\right) \\ &y=2x - 9 \end{align}[/latex]

Analysis of the Solution

We can use a graphing utility to graph the function and the tangent line. In so doing, we can observe the point of tangency at [latex]x=3[/latex] as shown in Figure 19.