An object is launched with a velocity at 64 feet per second from a platform 30 feet high. the function H(t)= -16t2+64t+30 repres
ents the function where T is time in seconds, and H is height in feet. what is the maximum height of the object ever reaches? a) 92 ft b) 94 ft c) 2 ft d) 94 ft
To find the maximum or minimum value of a function, we can find the derivative of the function, set it equal to 0, and solve for the critical points.
H'(t) = -32t + 64
Now find the critical numbers:
-32t + 64 = 0
-32t = -64
t = 2 seconds
Since H(t) has a negative leading coefficient, we know that it opens downward. This means that the critical point is a maximum value rather than a minimum. If we weren't sure, we could check by plugging in a value for t slightly less and slighter greater than t=2 into H'(t):
H'(1) = 32
H'(3) = -32
As you can see, the rate of change of the object's height goes from increasing to decreasing, meaning the critical point at t=2 is a maximum.
Answer: First, you identify how many rises and runs the line is from each point on a graph (intersection). Then you take the y's and the x's and do y2-y1/x2-x1 to get the slope!
