A ball is thrown straight up from a height of 3 ft with a speed of 32ft/s. It’s height above the ground after x seconds is given by the quadratic function y=-16x^2+32x+3 Explain the steps you would use to determine the path of the ball in terms of a transformation of the graph of y=X^2.

My first steps is to enter the equation into desmos and see what the coordinates are which is (1, 19) 1 to the left and 19 up. The function is narrow and goes below the x axis.

Step-by-step explanation: Got it right (btw you can copy my answer and put it on edge) that way our answers aren't all the same :) bye!!

on a coordinate grid, point a is at (−4.0, −2.4) and point b is at (4.0, −2.4). point b is a reflection of point a across the –axis. input either a lowercase x or i think is the right awnser

step-by-step explanation:

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)