An object moves along the x-axis. Its position is given by the equation y( X = 4t^2 - 41t+ 78 \). Find the position of the object when it changes direction

The object changes direction when its velocity changes sign. You can get the velocity function by differentiating the position function with respect to time t :

x(t) = 4t ² - 41t + 78

→   v(t) = dx(t)/dt = 8t - 41

Solve v(t) = 0:

8t - 41 = 0

8t = 41

t = 41/8 = 5.125

Just to confirm that the velocity indeed changes sign:

• Pick any time before this one to check the sign of v :

v (0) = 8•0 - 41 = -41 < 0

• Pick any time after and check the sign again:

v (6) = 8•6 - 41 = 7 > 0

Now just find the position at this time:

x (5.125) = -433/16 = -27.0625

which means the object is 27.0625 units on the negative x-axis.

You can also do this without calculus by completing the square in the position function:

4t ² - 41t + 78 = 4 (t ² - 41/4 t ) + 78

… = 4 (t ² - 2• 41/8 t + (41/8)² - (41/8)²) + 78

… = 4 (t ² - 2• 41/8 t + (41/8)²) - 4•1681/64 + 78

… = 4 (t - 41/8)² - 433/16

which describes a parabola that opens upward. When t = 41/8 = 5.125, the quadratic term vanishes and the turning point of the parabola occurs at a position of -433/16 units.

there are no correct statementd on the list of choices you provided.