Ahovercraft 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?

5 seconds

Step-by-step explanation:

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

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

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )