Xintercept (125,0)
Yintercept (0,275)
Answer:
The answer in the procedure
Step-by-step explanation:
we know that
The rule of the reflection of a point over the y-axis is equal to
A(x,y) ----->A'(-x,y)
That means -----> The x-coordinate of the image is equal to the x-coordinate of the pre-image multiplied by -1 and the y-coordinate of both points (pre-image and image) is the same
so
A(3,-1) ------> A'(-3,-1)
The distance from A to the y-axis is equal to the distance from A' to the y-axis (is equidistant)
therefore
To reflect a point over the y-axis
Construct a line from A perpendicular to the y-axis, determine the distance from A to the y-axis along this perpendicular line, find a new point on the other side of the y-axis that is equidistant from the y-axis
We take the equation <span>d = -16t^2+12t</span> and subtract d from both sides to get
0<span> = -16t^2+12t - d
We apply the quadratic formula to solve for t. With a = -16, b = 12, c = -d, we have
t = [ -(12) </span><span>± √( 12^2 - 4(-16)(-d) ) ] / [2 * -16]</span>
= [- 12 ± √(144-64d) ] / (-32)
= [- 12 ± √16(9-4d)] / (-32)
= [- 12 ± 4√(9-4d)] / (-32)
= 3/8 ±√(9-4d) / 8
The answer to your question is t = 3/8 ±√(9-4d) / 8
