Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
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 \(y = 4x + 23\).
Show All Steps Hide All Steps
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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{{5 - \sqrt {34} }}{3} = - 0.276984 \hspace{0.5in} x = \frac{{5 + \sqrt {34} }}{3} = 3.61032}}\]