10. Derivatives and Tangent Lines

b. Tangent Lines

The primary application of a derivative is to find the equation of the tangent line at a point, $x=a$, on the graph of a function, $y=f(x)$. The most general line has the standard equation: $Ax+By=C.$ The line is vertical if $B=0$ and non-vertical if $B\ne0$. Assuming it is not vertical, we can solve for $y$ and put the equation into slope-intercept form: $y=mx+b \qquad (1)$ where $m$ is the slope and $b$ is the $y$-intercept.

Now suppose we want to find the equation of the line tangent to $y=f(x)$ at $x=a$. We know the slope is $m=f'(a)$. So equation $(1)$ becomes: $y=f'(a)x+b \qquad (2)$ We know the line passes through the point $(x,y)=(a,f(a))$. So equation $(2)$ tell us: $f(a)=f'(a)a+b$ We solve for $b$ and substitute back into $(2)$: $\begin{aligned} b&=f(a)-f'(a)a \\[5pt] y&=f'(a)x+f(a)-f'(a)a\\ &=f(a)+f'(a)(x-a) \end{aligned}$ Using this formula for $b$, equation $(2)$ becomes: $\begin{aligned} y&=f'(a)x+f(a)-f'(a)a\\ &=f(a)+f'(a)(x-a) \end{aligned}$ which is the equation for the tangent line. We define the formula on the right to be the tangent function: $f_{\tan}(x)=f(a)+f'(a)(x-a)$ so that the equation of the tangent line is $y=f_{\tan}(x)$.

The equation of a tangent line can also be derived from the point-slope equation for a line. The slope is $m=\dfrac{y-y_o}{x-x_o}$ where $m$ is the slope, $(x_o,y_o)$ is some point on the line and $(x,y)$ is a general point on the line. Solving for $y$, the point-slope equation for the line is: $y=y_o+m(x-x_o)$ For the tangent line to $f(x)$ at $x=a$, the slope is $m=f'(a)$ and $(x_o,y_o)=(a,f(a)$. So the point-slope equation becomes $y=f(a)+f'(a)(x-a)$

Compute the tangent line to $y=6x^2-x^3$ at $x=3$. Then find its $y$-intercept. Its slope was found in an exercise on the previous page.

The solution is straightforward: find the required information and plug into the formula for the equation of the tangent line. We identify $f(x)=6x^2-x^3$ and $a=3$. Then: $f(3)=54-27=27$ and from the previous page: $f'(3)=9$ So the equation of the tangent line is: $\begin{aligned} y&=f(a)+f'(a)(x-a) \\ &=f(3)+f'(3)(x-3) \\ &=27+9(x-3) \\ &=9x \end{aligned}$ We identify the $y$-intercept as $b=0$.