A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. y= -16x² +198x+148​

x=0))

Step-by-step explanation:

S we want to find when the rocket will hit the ground (height=0 also y=0) so let setup the equation

-16x2+148x+128=0 (we can divide out a -4 so let's do that)

4x2-37x-32=0, now you can use the quadratic formula)

(37±sqrt(372-4(4)(-32))/(2*4)=x, x=10.046 & x=-.796 (we want the positive number because we want the time AFTER the launch (which is where x=0))

i think it is x = 1.2

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .