Section 3.3 : Differentiation Formulas
20. Determine where, if anywhere, the tangent line to \(f\left( x \right) = {x^3} - 5{x^2} + x\) is parallel to the line \(y = 4x + 23\).
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Start SolutionThe first thing that we’ll need of course is the slope of the tangent line. So, all we need to do is take the derivative of the function.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( x \right) = 3{x^2} - 10x + 1}}\]
Hint : What is the relationship between the slope of two parallel lines?
Two lines that are parallel will have the same slope and so all we need to do is determine where the slope of the tangent line will be 4, the slope of the given line. In other words, we’ll need to solve,
\[f'\left( x \right) = 4\hspace{0.25in} \to \hspace{0.25in}3{x^2} - 10x + 1 = 4\hspace{0.25in} \to \hspace{0.25in}3{x^2} - 10x - 3 = 0\]This quadratic doesn’t factor and so a quick use of the quadratic formula will solve this for us.
\[x = \frac{{10 \pm \sqrt {136} }}{6} = \frac{{10 \pm 2\sqrt {34} }}{6} = \frac{{5 \pm \sqrt {34} }}{3}\]So, the tangent line will be parallel to \(y = 4x + 23\) at,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{{5 - \sqrt {34} }}{3} = - 0.276984 \hspace{0.5in} x = \frac{{5 + \sqrt {34} }}{3} = 3.61032}}\]