Question 746086: I need help solving this Quadratic equation problem. I would appreciate it... If someone who may help walk me through steps to solving this problem. I need to learn this for my final that's coming up.
[Travel Speed (mph), Braking Distance (ft)]
[20(mph), 20 (ft)]
[40(mph), 105 (ft)]
[60(mph), 300(ft)]
a)Find a quadratic function that fits the data on the right.
b) Use the function to estimate the braking distance of a car that travels at 50 mph.
c) Does it make sense to use this function when speeds are less than 10 mph? Why or why not?
y=ax^2+bx+c
[20(mph), 20 (ft)] (20)=a(20)^2+b(20)+C = 20=400a+20b+C
[40(mph), 105 (ft)](105)=a(40)^2+b(40)+C = 105=1600a+40b+C
[60(mph), 300(ft)] (300)=a(60)^2+b(60)+C = 300=3600a+60b+C
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I need help solving this Quadratic equation problem. I would appreciate it... If someone who may help walk me through steps to solving this problem. I need to learn this for my final that's coming up.
[Travel Speed (mph), Braking Distance (ft)]
[20(mph), 20 (ft)]
[40(mph), 105 (ft)]
[60(mph), 300(ft)]
a)Find a quadratic function that fits the data on the right.
b) Use the function to estimate the braking distance of a car that travels at 50 mph.
c) Does it make sense to use this function when speeds are less than 10 mph? Why or why not?
y=ax^2+bx+c
[20(mph), 20 (ft)] (20)=a(20)^2+b(20)+C = 20=400a+20b+C
[40(mph), 105 (ft)](105)=a(40)^2+b(40)+C = 105=1600a+40b+C
[60(mph), 300(ft)] (300)=a(60)^2+b(60)+C = 300=3600a+60b+C
------
You now have 3 equations with three variables.
I used the matrix function on my TI-84 to solve that to get:
a = 0.138
b = -4
c = 45
--------------
f(x) = 0.138x^2-4x+45
f(50)= 190 ft.
f(10)= 18.8 ft
==================
Cheers,
Stan H.
==================
|
|
|
