Quadratic Functions

[vrfulgen]

http://i1202.photobucket.com/albums/bb380/vrfulgen/image_1358881903841286_zps76353195.jpg

the problem I'm having difficulty with is in the link. I'm having difficulty figuring out b. and e. I know you have to set the equation to 0 to find b. However I keep coming out with different answers, And I am not sure how to find the domain... I think it might be a common sense thing, maybe it's (0,infinity) or something, I have no idea.Any help will be greatly appreciated!! )

 

[kaitk]

If you have the means, try using a graphing calculator. The Ti-84, for instance, has various functions which may help you.

 

[HallsofIvy]

So you have the equation h= -4t^2+ 8t+ 10 which represents the height, in meters, above the water of a diver off a springboard t seconds after diving. The questions asked are
a) How high above the water is the springboard?
Yes, the diver is on the springboard when t= 0.. h(0)= 10 meters.

b) When does the diver enter the water?
Since h is the height above the water, when the diver enters the water, h= 0. So solve the equation -4t^2+ 8t+ 10= 0. You might want to simplify that by factoring out -2: -2(2t^2- 4t- 5)= 0 and dividing by -2 so that you have 2t^2- 4t- 5= 0. If you cannot not factor that easily, you could try "completing the square" or using the "quadratic formula". You probably are expected to be able to do at least one of those if you are given such a problem.

c) When does the diver reach her maximum height?
In my opinion, the best way (there are several) to do this is to complete the square as I suggested before. You can write the height function h= -2(2t^2-4t- 5)= -4(t^2- 2t- 5/2) as h= -4((t- a)^2- b)= -4(t- a)^2+ 4b for some numbers a and b. That is the equation of a parabola, opening downward, with its vertex, and so highest point, at (a, 4b). That is, if t= a, the "-4(t-a)^2" part is 0, leaving h= 4b. If a is any other number, (t- a)^2 is positive so -4(t-a)^2 is negative and -4(t- a)^2+ 4b is less than 4b.

I have a suspicion that this problem is in a section on completing the square!

 

[MarkFL]

Another approach to part c) is to locate the axis of symmetry, as this is where the vertex will be.

For the general quadratic \(\displaystyle f(x)=ax^2+bx+c\), the axis of symmetry is at \(\displaystyle x=-\dfrac{b}{2a}\).

 

[vrfulgen]

Thank you, everyone.

Yes it is a section on completing the square, finding the vertex and such.

But again, thanks I REALLY appreciate it!

-Victoria