what are the x-intercepts, the extent of the graph and the asymptotes for x^2/16-y^2/25 = 1

3 answers

What is wrong with just plugging in a zero for the y to get the x intercept? In this case it is 4, the sqrt of 16. What is meant by extent? Do you have a graph to which you are supposed to refer? If not, draw your own.
This time you have a hyperbola with
a = 4, b= 5
c^2 = 16 + 25 = 41
c= √41

Draw a rectangular box with centre at (0,0) and corners (±4,±5)
the asymptotes are the extended diagonals of that box.
the slopes will be 5/4 and -5/4, since it passes through the origin,
the equations of the asymptotes are
y = (5/4)x and y = (-5/4)x

Here is a webpage that sort of follows my method,
http://www.saskschools.ca/curr_content/mathc30/Unit5a/lesson4b_hyperbola.htm

Look about 1/2 way down the page
herbgeorge,
I don't have clue what is meant by "extent of the graph" either. I am trying to help my son with homework and there is no mention in the book about what it means or how to do it; it only ask for the correct answer. Yes, I do have a graph. I apologize that I don't know enough about algebra to understand it and all the terms.