Question

A rock is lanched from a cannon. It's height, h(x), can be represented by a quadratic function in terms of time, x in seconds.

After 1 second, the rock is 121 feet in the air;
After 2 seconds, it is 224 feel in the air.

Complete the height function, h(x), for this situation.

h(x) =

I'm so confused.. Help and please explain >.<

@thomaster

Answer 1

a quadratic function is in a form \[h(x)=ax^2+bx+c\]and the graph we expect is a parabola open downward, the symmetry is vertical.

Answer 2

Two points are already given that satisfies the quadratic function, they are (1,121) and (2,224)...

Answer 3

at time x=0, there is no h(x) yet and so h(0)=0...

Answer 4

Right, I got the function, however how do you get the answer? That's what I'm struggling with =/.

Answer 5

we can solve c right away at h(0)=0, it comes out that c=0

Answer 6

Okay.

Answer 7

for a and b, we have two points that we can substitute, so for the point (1,121) we can have \[121=a(1)^2+b(1)\]\[121=a+b~~~~(1)\]

Answer 8

we do the same for the second point (2,224)... do you think you can continue starting from here and solve 2 equations simultaneously to solve a and b?

Answer 9

Second point, what do you mean? O_o We aren't graphing??

Answer 10

as I said already, the quadratic function's graph is a parabola, the given situation in your problem serves as a point that satisfies the function... we have substitute the 1st point... and that is x=1 sec where h(1)=121 feet...

Answer 11

Ok

Answer 12

and for the second point, x=2 seconds, the resulting h(2)=224 feet, substitute this again, we have \[224=a(2)^2+b(2)\]\[224=4a+2b\]multiply both sides by \(\frac{1}{2}\)\[\frac{1}{\cancel{2}}\cancel{224}112=(2a+b)\cancel{2}\frac{1}{\cancel{2}}\]\[112=2a+b~~~~(2)\]

Answer 13

we now have the 2 equations, to solve for a and b \[121=a+b~~~~~(1)\]\[112=2a+b~~~~(2)\]

Answer 14

Ok.

Answer 15

(2)-(1)\[112=2a+\cancel{b}\]\[-121=-a-\cancel{b}\]\[112-121=2a-a\]\[-9=a~or~a=-9\]

Answer 16

using (1) to solve for b \[121=a+b\]\[121=-9+b\]\[b=130\]

Answer 17

there we can write now our expression for h(x) with a=-9, b=130, and c=0 \[h(x)=-9x^2+130x\]

Answer 18

... that finish the solution... and yet the above line indicates final answer... :)