I am taking a chance at this one.
See if you agree or not.
(0,0) is the initial starting point
(55, 0) is the ending point
24.8 is the max height which occurs 1/2 between 0 and 55
so that gives us another point
(27.5, 24.8)
I used the quadratic regression on the graphing calculator to get an equation.
im not sure how to do this question
A student fires a model rocket into the air. The rocket reaches a maximum hieght of 24.8 meters and lands 55 meters from the student. (assume parabola and no wind)
a. list three points of trajectory(coordinate pairs)
b. determine the regression equation that best fits the trajectory using your graphing calculator
4 answers
ok thank you i will try that i have no idea how to use my calculator so i will try and do what you said.
I put the numbers in using the stat function of the TI-83.
I hit STAT and edit to put in the numbers in L1 and L2.
Then I did Stat -Calc and I chose Quadreg and L1 L2 from keyboard and enter.
I hit STAT and edit to put in the numbers in L1 and L2.
Then I did Stat -Calc and I chose Quadreg and L1 L2 from keyboard and enter.
yes that worked in the calulator and i had an idea that those would be the points.
using the graphing calulator how do you find the regression equation? could you walk me through it? so i can do it on the calculator ?
using the graphing calulator how do you find the regression equation? could you walk me through it? so i can do it on the calculator ?