Question

Task 4:

A Jesus Christ lizard is jumping across the water in search of
food. The equation h = -12t2 + 6t models the lizard's height
in feet above the water t seconds after he jumps.
A: How long after jumping is he back on the water?
0,3
22
B: How high is each jump?
-12(0.251
075 teet
C: How long does it take to get to
his highest point? 0.25

Answer 1

A: To determine when the lizard is back on the water, we need to find the time when the height (h) is equal to 0. So we set the equation -12t^2 + 6t = 0 and solve for t.

-12t^2 + 6t = 0

Factor out common terms:

-6t(2t - 1) = 0

Set each factor equal to 0:

-6t = 0 or 2t - 1 = 0

Solving each equation:

-6t = 0 --> t = 0

2t - 1 = 0 --> 2t = 1 --> t = 1/2

So the lizard is back on the water at t = 0 seconds and t = 1/2 seconds.

B: The height of each jump can be determined by substituting the time (t) values into the equation h = -12t^2 + 6t.

For t = 0 seconds:

h = -12(0)^2 + 6(0)

h = 0

For t = 1/2 seconds:

h = -12(1/2)^2 + 6(1/2)

h = -12(1/4) + 6/2

h = -3 + 3

h = 0

So each jump has a height of 0 feet.

C: To find the time it takes to reach the highest point, we need to find the vertex of the parabolic equation -12t^2 + 6t. The time at the vertex represents the highest point.

The formula for the x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is given by -b/(2a). In this case, a = -12 and b = 6.

t = -6/(2(-12))

t = -6/(-24)

t = 1/4

So it takes 1/4 seconds to reach the highest point.

Therefore, the answers are:

A: The lizard is back on the water at t = 0 seconds and t = 1/2 seconds.

B: Each jump has a height of 0 feet.

C: It takes 1/4 seconds to reach the highest point.