Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

Can anyone help me with a math project? Or at least explain how to do it.

OpenStudy (ash2326):

Refer this , it may help http://mathematicsprojects.blogspot.in/

OpenStudy (anonymous):

@ash2326 that doesn't help cause my project isnt on there

OpenStudy (anonymous):

can anyone help me?

OpenStudy (anonymous):

what do you mean by that?

OpenStudy (anonymous):

my project involves vertex form and graphing based on a football statement

OpenStudy (hba):

So you have a question or Assignment?

OpenStudy (anonymous):

i dont understand how to do the project i just want help explaining how to do it that is it

OpenStudy (anonymous):

what kind of a project?

OpenStudy (anonymous):

a football player kicked a 40-yard punt. The path of the ball was modeled by y = -0.0375x^2 + 1.5x, where x and y are measured in yards. What was the maximum height of the ball? The player kicked the football toward midfield from the 18 yard line. Over which yard line was the ball when it was its maximum height?

OpenStudy (anonymous):

and you have to draw a picture of the problem then use vertex formula to write the equation in graphing form then graph it then find the x-intercept and the y-intercepts of the graph

OpenStudy (anonymous):

max height is at the vertex

OpenStudy (anonymous):

yes but how do i get that?

OpenStudy (anonymous):

here is a picture http://www.wolframalpha.com/input/?i=-0.0375x^2+%2B+1.5x%2C+domain+0..40

OpenStudy (anonymous):

ok thanks @satellite73

OpenStudy (anonymous):

first coordinate of the vertex is \(-\frac{b}{2a}\) which in your case it \(-\frac{1.5}{2\times .0375}=20\)

OpenStudy (anonymous):

if you replace \(x\) by \(20\) you get \(y=15\) so the max height is 15 feet up

OpenStudy (anonymous):

ok thanks

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!