The object changes direction when its velocity changes sign. You can get the velocity function by differentiating the position function with respect to time <em>t</em> :
<em>x(t)</em> = 4<em>t</em> ² - 41<em>t</em> + 78
→ <em>v(t)</em> = d<em>x(t)</em>/d<em>t</em> = 8<em>t</em> - 41
Solve <em>v(t)</em> = 0:
8<em>t</em> - 41 = 0
8<em>t</em> = 41
<em>t</em> = 41/8 = 5.125
Just to confirm that the velocity indeed changes sign:
• Pick any time before this one to check the sign of <em>v</em> :
<em>v</em> (0) = 8•0 - 41 = -41 < 0
• Pick any time after and check the sign again:
<em>v</em> (6) = 8•6 - 41 = 7 > 0
Now just find the position at this time:
<em>x</em> (5.125) = -433/16 = -27.0625
which means the object is 27.0625 units on the negative <em>x</em>-axis.
You can also do this without calculus by completing the square in the position function:
4<em>t</em> ² - 41<em>t</em> + 78 = 4 (<em>t</em> ² - 41/4 <em>t</em> ) + 78
… = 4 (<em>t</em> ² - 2• 41/8 <em>t</em> + (41/8)² - (41/8)²) + 78
… = 4 (<em>t</em> ² - 2• 41/8 <em>t</em> + (41/8)²) - 4•1681/64 + 78
… = 4 (<em>t</em> - 41/8)² - 433/16
which describes a parabola that opens upward. When <em>t</em> = 41/8 = 5.125, the quadratic term vanishes and the turning point of the parabola occurs at a position of -433/16 units.
