what are the x-intercepts, the extent of the graph and the asymptotes for x^2/16-y^2/25 = 1
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.
To find the x-intercepts, extent of the graph, and asymptotes of the equation x^2/16 - y^2/25 = 1, we can use the properties of hyperbolas.
To start, let's rearrange the equation to the standard form of a hyperbola:
(x^2/16) - (y^2/25) = 1
The equation tells us that the denominator of the x-term is 16, while the denominator of the y-term is 25. Taking the square root of these denominators will give us the values necessary to find the vertices and the endpoints of the conjugate axis.
The equation represents a hyperbola with the x-axis as the transverse axis. This means that the x-term has a positive coefficient, while the y-term has a negative coefficient.
To find the x-intercepts, we set y = 0 and solve for x:
(x^2/16) - (0/25) = 1
(x^2/16) = 1
x^2 = 16
Taking the square root of both sides, we get:
x = ±4
So the x-intercepts are x = -4 and x = 4.
To find the extent of the graph, we look at the denominators of the x and y terms. In this case, the denominators are 16 and 25, respectively. Taking the square root of these denominators, we get 4 and 5.
The extent of the graph along the x-axis is determined by the value under the x^2 term, so the graph will extend 4 units to the left and 4 units to the right of the origin along the x-axis.
The extent of the graph along the y-axis is determined by the value under the y^2 term, so the graph will extend 5 units upward and 5 units downward from the origin along the y-axis.
Next, let's find the asymptotes of the hyperbola. For this equation, since we have a horizontal transverse axis (the x-axis), the asymptotes will be given by the equations:
y = ± (b/a) * x + k
Where b is the square root of the value under the y^2 term (in this case, 5), and a is the square root of the value under the x^2 term (in this case, 4). k represents the y-coordinate of the center, which is 0 since the hyperbola is centered at the origin.
Therefore, the asymptotes are:
y = (5/4) * x
y = -(5/4) * x
To summarize:
- The x-intercepts are x = -4 and x = 4.
- The extent of the graph is 4 units to the left and right along the x-axis, and 5 units up and down along the y-axis.
- The asymptotes of the hyperbola are y = (5/4) * x and y = -(5/4) * x.