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: \[ f(a)=f'(a)a+b \] or \[ b=f(a)-f'(a)a \] 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 \(y=f_{\tan}(x)\).

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 \(b=0\).