A hovercraft takes off from a platform. Its height (in meters), xx seconds after takeoff, is modeled by: h(x)=-2x^2+20x+48h(x)=?
2x 2 +20x+48 How many seconds after takeoff will the hovercraft reach its maximum height?
2 answers:
Answer: 5 seconds
<u>Step-by-step explanation:</u>
Maximum height is the y-value of the vertex. The number of seconds at maximum height is the x-value (aka axis of symmetry).
h(x) = -2x² + 20x + 48
a=-2 b=20 c=48
Axis of symmetry: x = 
= 
= 
= 5
Answer:
5seconds
Step-by-step explanation:
Given the height of over craft given as h(x)=-2x^2+20x+48, at maximum height, the velocity of the object will be zero I.e v = d{h(x)}/dx = 0
d{h(x)}/dx = -4x + 20
If d{h(x)}/dx = 0, then
-4x + 20 = 0
Moving 20 to the other side;
-4x = 0-20
-4x = -20
x =-20/-4
x = 5
This shows that the time it takes hovercraft to reach its maximum height is 5seconds
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